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int64 1
7.14k
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1,801
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21
| 5
|
A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
|
Through observation, we can reasonably assume that each of the triangles on this shape is a right triangle. Since each side length of the hexagons is $1$ , the hypotenuse of the triangles would be $\sqrt2$ . Now we know the side lengths of the octagon whose area we are solving for. The octagon can be broken into nine pieces. We have four triangles whose side lengths are 1, and their hypotenuse is a side whose length is $\sqrt2$ . Next, we have $5$ $1$ by $1$ squares. The triangles each have an area of $\frac{1}{2}$ , and the squares each have an area of $1$
Then, we add these up, so we get $\boxed{7}.$
| 7
|
1,802
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22
| 1
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement [asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("1", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("2", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("3", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("4", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("5", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("6", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("7", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("8", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("9", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("10", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("11", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("12", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("13", (37,1.5)); [/asy] in which the first pass consists of all $13$ cards.
Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{8178}.\]
| 178
|
1,803
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22
| 2
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ .
Also note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.
To construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ .
Let's do casework on which slot $n$ goes into to get a general idea for how the problem works.
Case 1: With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\binom{n}{n-1}$ possibilities.
Case 2: With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\binom{n+1}{n-1}$ possibilities.
I think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: \[\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\] for any $n$
Hmmm ... where have we seen this before?
We use wishful thinking to add a term of $\binom{n-1}{n-1}$ \[\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\]
This is just the hockey stick identity! Applying it, this expression is equal to $\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \[\binom{13}{n}-1\]
Finally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:
\[\sum_{n=0}^{13} \binom{13}{n}-1 \implies \sum_{n=0}^{13} \binom{13}{n} - \sum_{n=0}^{13} 1\] \[\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{8178}.\]
| 178
|
1,804
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22
| 3
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$
If $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.
If $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{1}$ such arrangements.
If $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{k-1}$ such arrangements.
So we can write \[A_n = A_{n-1} + \binom{n-1}{1} + \binom{n-1}{2} + \cdots + \binom{n-1}{n-1}\] which can be simplified to \[A_n = A_{n-1} + 2^{n-1} - 1\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \[A_n - A_{n-1} = 2^{n-1} - 1\] \[A_{n-1} - A_{n-2} = 2^{n-2} - 1\] \[\cdots\] \[A_2 - A_{1} = 2^{1} - 1\] to get \[A_n - A_1 = \sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\] since $A_1=0$ , we have \[A_n = 2^n - n - 1\] and $A_{13} = 2^{13} - 14 = \boxed{8178}$
| 178
|
1,805
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22
| 4
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \boxed{8178}.$ ~eibc
| 178
|
1,806
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22
| 5
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
Notice that for each card "position", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and thus all cards will be picked up on the first pass instead). This can happen in 14 ways, so our answer is $2^{13}-14=\boxed{8178}$
| 178
|
1,807
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24
| 1
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{1296}$ such strings.
| 296
|
1,808
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24
| 2
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$
| 296
|
1,809
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24
| 3
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$
By solving this recursive equation, for $q = 1$ and $p \geq q$ , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}
For $q = 2$ and $p \geq q$ , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}
For $q = 3$ and $p \geq q$ , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}
For $q = 4$ and $p = 5$ , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{1296}
| 296
|
1,810
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24
| 4
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$
| 296
|
1,811
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24
| 5
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).
These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{1296}$
| 296
|
1,812
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25
| 1
|
Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
|
Let $r$ be the number of lattice points on the side length of square $R$ $s$ be the number of lattice points on the side length of square $S$ , and $t$ be the number of lattice points on the side length of square $T$ . Note that the actual lengths of the side lengths are the number of lattice points minus $1$ , so we can work in terms of $r, s, t$ and subtract $3$ to get the actual answer at the end. Furthermore, note that the number of lattice points inside a rectangular region is equal to the number of lattice points in its width times the number of lattice points along its length.
Using this fact, the number of lattice points in $R$ is $r^2$ , the number of lattice points in $S$ is $s^2$ , and the number of lattice points in $T$ is $t^2$
Now, by the first condition, we have \[r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s \quad \quad \quad \quad \quad (1)\]
The second condition, the number of lattice points contained in $T$ is a fourth of the number of lattice points contained in $R \cup S$ . The number of lattice points in $R \cup S$ is equal to the sum of the lattice points in their individually bounded regions, but the lattice points along the y-axis for the full length of square $S$ is shared by both of them, so we need to subtract that out.
In all, this condition yields us $t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot\left(\frac{9}{4}\cdot s^2 + s^2 - s \right)$ $\implies t^2=\frac{1}{4}\cdot\frac{13s^2-4s}{4} \implies 16t^2= s(13s-4)$
Note from $(1)$ that $s$ is a multiple of $2$ . We can write $s=2j$ and substitute: $16t^2=2j(26j-4) \implies 4t^2=j(13j-2)$ .
Note that $j$ must be divisible by two for the product to be divisible by 4. Thus we make another substitution, $j=2k$ \[4t^2=2k(26k-2) \implies t^2 = k(13k-1) \quad \quad \quad \quad \quad (2)\]
Finally we look at the last condition; that the fraction of the lattice points inside $S$ that are inside $S \cap T$ is $27$ times the fraction of lattice points inside $r$ that are inside $R \cap T$
Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$ , and $y$ be the number of lattice points along the bottom of the the rectangle formed by $R \cap T$
Therefore, the number of lattice points in $S\cap T$ is $xt$ and the number of lattice points in $R \cap T$ is $yt$
Thus by this condition, $\frac{xt}{s^2} = 27 \cdot \frac{yt}{r^2} \implies \frac{x}{s^2} = 27 \cdot \frac{y}{\frac{9}{4}\cdot s^2} \implies x= 12y$
Finally, notice that $t=x+y-1=12y+y-1$ (subtracting overlap), and so we have \[t=13y-1 \quad \quad \quad \quad \quad (3)\]
Now notice that by $(3)$ $t\equiv -1 \pmod{13}\implies t^2 \equiv 1 \pmod{13}$
However, by $(2)$ $t^2 \equiv k \cdot -1 \pmod{13}$ . Therefore, $-k \equiv 1 \pmod{13} \implies k \equiv -1 \pmod{13}$
Also, by $(2)$ , we know $k$ must be a perfect square since $k$ is relatively prime to $13k-1$ (Euclids algorithm) and the two must multiply to a perfect square. Hence we know two conditions on $k$ , and we can now guess and check to find the smallest that satisfies both.
We check $k=12$ first since its one less than a multiple of $13$ , but this does not work. Next, we have $k=25$ which works because $25$ is a perfect square. Thus we have found the smallest $k$ , and therefore the smallest $r, s, t$
Now we just work backwards: $j= 2k = 50$ and $s=2j=100$ . Then $r=\frac{3}{2}\cdot 100 = 150$ . Finally, from $(2)$ $t^2=25(13\cdot25-1) \implies t^2 = 25 \cdot 324 \implies t=5\cdot 18=90$
Finally, the sum of each square’s side lengths is $r+s+t-3=340-3=337=\boxed{337}$
| 337
|
1,813
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25
| 2
|
Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
|
Notice that each answer choice has a different residue mod $13$ . Therefore, we can just find the residue of $r+s+t-3$ mod $13$ and find the unique answer choice that fits, without actually finding $r, s, t$
From Solution 1, we have $16t^2 = s(13s-4)$ from the second condition. From the third condition, $t\equiv -1 \pmod{13} \implies t^2 \equiv 1 \pmod{13}$ .
Substituting, we get $16 \cdot 1 \equiv s \cdot -4 \pmod{13}$ . Therefore, $s \equiv -4 \pmod{13}$ .
From the first condition, we have $r=\frac{3}{2} \cdot s$ , so $r \equiv -6 \pmod{13}$
Therefore $r+s+t \equiv -6-4-1 \equiv -11 \equiv 2 \pmod {13}$
We want to find $r+s+t-3$ , so our answer will have a remainder of $-1$ when divided by $13$
We divide $340$ by $13$ and find that the remainder is $2$ . Therefore the answer that will give us a remainder of $-1$ will be $340-3=337=\boxed{337}$
| 337
|
1,814
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25
| 3
|
Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
|
Solution: Let $r$ $s$ $t$ be the edge length of square $R$ $S$ , and $T$ respectively. Then we have \[(r+1)^2=\dfrac{9}{4}(s+1)^2\ \ \ \ \ (t+1)^2=\dfrac{1}{4}((s+1)^2+(r+1)^2-(s+1))\] Therefore \[r=\dfrac{3s+1}{2}\ \ \ \ \ t=\dfrac{1}{4}\sqrt{(s+1)(13s+9)}-1\] Therefore \[r+s+t=\dfrac{3s+1}{2}+s+\dfrac{1}{4}\sqrt{(s+1)(13s+9)}-1\] \[\approx\dfrac{5}{2}s+\dfrac{\sqrt{13}}{4}s-\dfrac{1}{2}\approx 3.4\cdot s\]
Given that average of the answer choices is around $340$ , therefore $s\approx 100$ . Since $t$ is an integer, therefore $(s+1)(13s+9)$ must be a perfect square divisible by 16. Plugging in $s=99$ $t=89$ and $r=149$ . Therefore $r+s+t=99+89+149=337$ . So the answer is $\boxed{337}$
| 337
|
1,815
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_1
| 1
|
Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$
|
We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM
| 2
|
1,816
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_1
| 2
|
Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$
|
Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$
| 2
|
1,817
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_2
| 1
|
In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]
\[\textbf{Figure redrawn to scale.}\]
$AD = AP + PD = 3 + 2 = 5$
$ABCD$ is a rhombus, so $AB = AD = 5$
$\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$
The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{20}$
| 20
|
1,818
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_2
| 2
|
In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]
The diagram is from as Solution 1, but a line is constructed at $BD$
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that $\angle ABD \cong \angle BDC$ , by the Alternate Interior Angles Theorem.
By SAS Congruence, we get $\triangle ABD \cong \triangle BDC$
Since $AP=3$ and $AB=5$ , we know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle, as stated in Solution 1.
The area of $\triangle ABD$ would be $10$ , since the area of the triangle is $\frac{bh}{2}$
Since we know that $\triangle ABD \cong \triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$ , so we can double the area of $\triangle ADB$ to get $10 \cdot 2 = \boxed{20}$
| 20
|
1,819
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_3
| 1
|
How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$
|
We use simple case work to solve this problem.
Case 1: even, even, even = $4 \cdot 5 \cdot 5 = 100$
Case 2: even, odd, odd = $4 \cdot 5 \cdot 5 = 100$
Case 3: odd, even, odd = $5 \cdot 5 \cdot 5 = 125$
Case 4: odd, odd, even = $5 \cdot 5 \cdot 5 = 125$
Simply sum up the cases to get your answer. $100 + 100 + 125 + 125 = \boxed{450}$
| 450
|
1,820
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_3
| 2
|
How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$
|
We will show that the answer is $450$ by proving a bijection between the three digit integers that have an even number of even digits and the three digit integers that have an odd number of even digits. For every even number with an odd number of even digits, we increment the number's last digit by $1$ , unless the last digit is $9$ , in which case it becomes $0$ . It is very easy to show that every number with an even number of even digits is mapped to every number with an odd number of even digits, and vice versa. Thus, the answer is half the number of three digit numbers, or $\boxed{450}$
| 450
|
1,821
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5
| 1
|
What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$
|
We apply the difference of squares to the denominator, and then regroup factors: \begin{align*} \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ &= \frac{\sqrt8}{\sqrt2} \\ &= \boxed{2} ~MRENTHUSIASM
| 2
|
1,822
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5
| 2
|
What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$
|
Since these numbers are fairly small, we can use brute force as follows: \[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} =\frac{\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7}}{\sqrt{\frac{8}{9}\cdot\frac{24}{25}\cdot\frac{48}{49}}} =\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} =\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{2}.\] ~not_slay
| 2
|
1,823
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5
| 3
|
What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$
|
This solution starts precisely like the one above. We simplify to get the following:
\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}}\]
But now, we can get a nice simplification as shown: \[\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{2^{10} \cdot 3^{2}}{3^2\cdot 5^2\cdot 7^2}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\frac{2^5 \cdot 3}{3\cdot5\cdot 7}} =\dfrac{4\cdot6\cdot8}{3\cdot5\cdot7} \hspace{0.05 in} \cdot \hspace{0.05 in}\dfrac{3\cdot5\cdot 7}{2^5 \cdot 3} =\dfrac{2^6\cdot 3}{2^5\cdot 3} = \boxed{2}.\]
| 2
|
1,824
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6
| 1
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
The $n$ th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.
| 0
|
1,825
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6
| 2
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.
| 0
|
1,826
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6
| 3
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{0}$ primes in the set.
| 0
|
1,827
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6
| 4
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$
| 0
|
1,828
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7
| 2
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4\cdot2=\boxed{8}$ values of $k$
| 8
|
1,829
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7
| 3
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $12$ $15$ triangle, a $12$ $16$ $20$ triangle, a $5$ $12$ $13$ triangle, and a $12$ $35$ $37$ triangle.
Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{8}$
| 8
|
1,830
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7
| 4
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$
| 8
|
1,831
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_8
| 1
|
Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$
|
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$
| 42
|
1,832
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_8
| 2
|
Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$
|
We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.
We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{42}$
| 42
|
1,833
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9
| 1
|
The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$
|
Note that $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$ , and therefore this sum is a telescoping sum, which is equivalent to $1 - \frac{1}{2022!}$ . Our answer is $1 + 2022 = \boxed{2023}$
| 23
|
1,834
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9
| 2
|
The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$
|
We add $\frac{1}{2022!}$ to the original expression \[\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}.\] This sum clearly telescopes, thus we end up with $\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1$ . Thus the original expression is equal to $1-\frac{1}{2022!}$ , and $1+2022=\boxed{2023}$
| 23
|
1,835
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9
| 4
|
The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$
|
Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$
Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{2023}.$
| 23
|
1,836
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9
| 5
|
The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$
|
Because the fractions get smaller, it is obvious that the answer is less than $1$ , so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$ . Because the last term consists of $2022!$ (and the year is $2022$ ) we can guess that $b=2022$ . Adding them yields $1+2022=\boxed{2023}$
| 23
|
1,837
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9
| 6
|
The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$
|
Knowing that the answer will be in the form $a-\frac{1}{b!}$ , we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$ . Setting these equal and multiplying by $n!$ , we get $n-1=An-B(n-1)!$ . Since $An$ is the only term with $n$ with degree $1$ , we can conclude that $A=1$ . This means that $B=\frac{1}{(n-1)!}$ . Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$ . This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$ . This means that our desired answer is $1+2022=\boxed{2023}.$
| 23
|
1,838
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_10
| 1
|
Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$
|
Let $M$ be the median. It follows that the two largest integers are both $M+2.$
Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{11}.$
| 11
|
1,839
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_10
| 2
|
Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$
|
We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
For answer choice $\textbf{(A)},$ the mode is $5,$ the median is $3,$ and the arithmetic mean is $1.$ However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of $1$ while having a mode of $5.$
Trying answer choice $\textbf{(B)},$ the mode is $7,$ the median is $5,$ and the arithmetic mean is $3.$ From the arithmetic mean, we know that all the numbers have to sum to $15.$ We know three of the numbers: $\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.$ This exceeds the sum of $15.$
Now we try answer choice $\textbf{(C)}.$ The mode is $9,$ the median is $7,$ and the arithmetic mean is $5.$ From the arithmetic mean, we know that the list sums to $25.$ Three of the numbers are $\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,$ which is exactly $25.$ However, our list needs positive integers, so this won't work.
Since we were really close on answer choice $\textbf{(C)},$ we can intuitively feel that the answer is probably going to be $\textbf{(D)}.$ We can confirm this by creating a list that satisfies the problem and choose $\textbf{(D)}: 1,3,9,11,11.$
So, our answer is $\boxed{11}.$
| 11
|
1,840
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12
| 1
|
A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$
|
Rolling a pair of fair $6$ -sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$
Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{4}.$
| 4
|
1,841
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12
| 2
|
A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$
|
Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$ , the first and third sum to $7$ , or the second and third sum to $7$ . There are $6$ ways for the first and second dice to sum to $7$ $6$ ways for the first and third to sum to $7$ , and $6$ ways for the second and third dice to sum to $7$ . However, we overcounted (but not by much) so we can assume that the answer is $\boxed{4}$
| 4
|
1,842
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12
| 3
|
A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$
|
We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\frac16$ , as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for $n$ rolls would be $\frac16$ $n$ . The smallest number that satisfies this is $\boxed{4}$
| 4
|
1,843
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13
| 1
|
The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$
|
Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$
Now let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$
Notice how if we add 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\boxed{16}$
| 16
|
1,844
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13
| 2
|
The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$
|
Let the two primes be $a$ and $b$ , with $a$ being the larger prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, \[a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549\] \[ab = 5183.\] Because we have $b = a+2$ $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \boxed{16}$
| 16
|
1,845
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13
| 3
|
The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$
|
Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$
By the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$
Plugging in $p-q=2$ and $p^{3}-q^{3}=31106$
$31106=2(p^{2}+pq+q^{2})$
Through the givens, we can see that $p \approx q$
Thus, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200$
Recall that $70^2=4900$ and $80^2=6400$ . It follows that our primes must be only marginally larger than $70$ , where we conveniently find $p=73, q=71$
The least prime greater than these two primes is $79 \implies 7 + 9 \implies \boxed{16}$
| 16
|
1,846
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13
| 4
|
The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$
|
Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \[6x^2 + 2 = 31106 \implies x^2 = 5184.\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger than $73$ is $79$ , which has a digit sum of $7 + 9 = \boxed{16}.$
| 16
|
1,847
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14
| 1
|
Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
|
Let $M$ be the largest number in $S$ .
We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups, such that the $i$ th group contains two numbers $i$ and $M-i$
Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$ , and the sum of numbers in each group above is equal to $S$ . Thus, each of the above $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups can have at most one number in $S$ .
Therefore, \begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13. \end{align*}
Next, we construct an instance of $S$ with $|S| = 13$ .
Let $S = \left\{ 13, 14, \ldots , 25 \right\}$ .
Thus, this set is feasible.
Therefore, the most number of elements in $S$ is $\boxed{13}$
| 13
|
1,848
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14
| 2
|
Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
|
We know that two odd numbers sum to an even number, so we can easily say that odd numbers $1-25$ can be included in the list, making for $13$ elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than $13$ , odd numbers as well as even numbers would have to be included in the list (since there are only $12$ even numbers from $1-25$ , and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, $\boxed{13}$ elements is the highest we can go.
| 13
|
1,849
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14
| 3
|
Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
|
The smallest sum of a number $a + b$ where $b \geq a$ is $a + a = 2a$ as we are using the smallest value of $b$ . Using this, we can say that if $12$ were an element of $S$ , then one of the sums (the smallest) would be $12 + 12 = 24 < 25$ . Thus $13$ must be the smallest element. So the largest amount of elements that could be in $S$ is the list of numbers from $13$ to $25$ as they all work. Because it is inclusive we have, $25 - 13 + 1 = \boxed{13}$
| 13
|
1,850
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14
| 4
|
Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
|
We construct a possible subset $S$ with $13$ elements by including all odd integers from $1$ to $25$ , inclusive. $S=\left\{ 1, 3, 5, \cdots , 25 \right\}$ . The sum of any $2$ elements is even, and thus cannot be an element of $S$
To show that $S$ cannot have more than $13$ elements, assume for sake of contradiction that $|S| \geq 14$ . Let $S=\left\{ x_1, x_2, \cdots , x_n \right\}$ where $n \geq 14$ and $x_1 < x_2 < \cdots < x_n$ . Because the sums of any $2$ (not necessarily distinct) elements do not appear in $S$ $x_1+x_i$ is not an element of $S$ for all $1 \leq i \leq n$ . So, $x_1, x_2, \cdots , x_n , x_1+x_1, x_1+x_2, \cdots , x_1+x_n$ are all distinct integers. Let these integers be elements of the set $T$ $|T|=2n$ , and because $n \geq 14$ $|T| \geq 28$ . But all elements of $T$ must be $\geq x_1$ and $\leq x_1+x_n \leq x_1+25$ , leaving only 26 possible values for the elements in $T$ . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.
Thus, $\boxed{13}$ is the maximum possible size of $S$
| 13
|
1,851
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14
| 5
|
Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$
|
We can start by building a list of the elements of $S$ ,and see if we can find a pattern. Let's start with $1$ . If we include 1, that means we cannot include $2$ (which is $1$ $1$ ), so we write the next valid number, $3$ . Similarly, we cannot include $4$ or $6$ , so we write $5$ . Proceeding, our list looks like this:
$S$ = { $1$ $3$ $5$ $7$ ....
We've found a pattern! It's easy understand why this pattern of odd numbers will continue, seeing as all the even numbers can be written as sums between odds. Calculating how many odd numbers fit into the range $1$ $25$ , we conclude that the answer is $\boxed{13}$
| 13
|
1,852
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_15
| 1
|
Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$
|
Suppose that the first number of the arithmetic sequence is $a$ . We will try to compute the value of $S_{n}$ . First, note that the sum of an arithmetic sequence is equal to the number of terms multiplied by the median of the sequence. The median of this sequence is equal to $a + n - 1$ . Thus, the value of $S_{n}$ is $n(a + n - 1) = n^2 + n(a - 1)$ . Then, \[\frac{S_{3n}}{S_{n}} = \frac{9n^2 + 3n(a - 1)}{n^2 + n(a - 1)} = 9 - \frac{6n(a-1)}{n^2 + n(a-1)}.\] Of course, for this value to be constant, $6n(a-1)$ must be $0$ for all values of $n$ , and thus $a = 1$ . Finally, we have $S_{20} = 20^2 = \boxed{400}$
| 400
|
1,853
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_15
| 2
|
Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$
|
Recall that the sum of the first $n$ odd numbers is $n^2$
Since $\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9$ , we have $S_n = 20^2 = \boxed{400}$
| 400
|
1,854
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18
| 1
|
Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$
|
Let $M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix},$ and $M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.$
We wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework:
There are $9+3=12$ ordered triples $(M_1,M_2,M_3).$
Similarly, for each of $M_1+M_3=M_2$ and $M_2+M_3=M_1,$ there are $12$ ordered triples $(M_1,M_2,M_3).$
In this subcase, we have $\boldsymbol{12\cdot3=36}$ ordered triples $\boldsymbol{(M_1,M_2,M_3).}$
Together, the answer is $8+168+126+36=\boxed{338}.$
| 338
|
1,855
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18
| 2
|
Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$
|
We will use complementary counting and do casework on the equations.
There are $8$ possible equations:
Equation 1: $0 = 0$
Equation 2: $x = 0$
Equation 3: $y = 0$
Equation 4: $z = 0$
Equation 5: $x + y = 0$
Equation 6: $x + z = 0$
Equation 7: $y + z = 0$
Equation 8: $x + y + z = 0$
We will continue to refer to the equations by their number on this list.
$8^3 = 512$ total systems. Note that no two equations by themselves can force $x = y = z = 0$ . Therefore no system with Equation 1 or with repeated equations can force $x = y = z = 0$
Case 1: Equation 8 ( $x + y + z = 0$ ) is present.
Case 1a: Equation 8, and two equations from $\{5, 6, 7\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{5, 6, 7\}$ and $3! = 6$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 3! = 18$
Case 1b: Equation 8, one equation from $\{5, 6, 7\}$ , and one equation from $\{2, 3, 4\}$
There are $\binom{3}{1} = 3$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let us choose Equation 7. Given $x + y + z = 0$ and $y + z = 0$ , we conclude that $x = 0$ . The third equation can be either $y = 0$ or $z = 0$ . There are $3!$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 2 \cdot 3! = 36$
Case 1c: Equation 8, and two equations from $\{2, 3, 4\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{2, 3, 4\}$ and $3! = 6$ ways to arrange each case. Each of these cases forces $x = y = z = 0$ $3 \cdot 3! = 18$ total options.
Case 2: Equation 8 is $\textbf{not}$ present, at least one equation from $\{5, 6, 7\}$ is present.
Case 2a: Equations $\{5, 6, 7\}$ are all present.
There are $3!$ ways to arrange the three equations. $6$ options.
Case 2b: Two equations from $\{5, 6, 7\}$ are present. One equation from $\{2, 3, 4\}$ is present.
There are $\binom{3}{2}$ ways to choose two equations from $\{5, 6, 7\}$ . WLOG let Equations 5 and 6 be in our system: $x + y = 0$ and $x + z = 0$ . Any equation from $\{2, 3, 4\}$ will force $x = y = z = 0$ . There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{2} \cdot \binom{3}{1} \cdot 3! = 54$
Case 2c: One equation from $\{5, 6, 7\}$ is present. Two equations from $\{2, 3, 4\}$ are present.
There are $\binom{3}{1}$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let Equation 5 ( $x + y = 0$ ) be present. One of the two equations from $\{2, 3, 4\}$ must be Equation 4, $z = 0$ , since it is the only equation that restricts $z$ . The last equation can be either 2 or 3. There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{1} \cdot \binom{2}{1} \cdot 3! = 36$
Case 3: Only equations $\{2, 3, 4\}$ are present.
There are $3!$ ways to arrange the three equations. $6$ options.
We add up the cases: $18 + 36 + 18 + 6 + 54 + 36 + 6 = 174$ total systems force $x = y = z = 0$ . Thus $512 - 174 = \boxed{338}$ do not.
| 338
|
1,856
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18
| 3
|
Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$
|
The total number of possible systems is $2^9 = 512$ , with $8$ possible sets of coefficients per equation. We will use complementary counting to find the number of systems which only have the solution $(0, 0, 0)$ and subtract that from the total. Similar to what is observed in Solution 2, if any equation is repeated or $0x + 0y + 0z = 0$ , there will only be two or fewer equations for three variables, making one unique solution impossible. Therefore, we must choose $3$ different equations from $7$ possible ones, giving $7 \cdot 6 \cdot 5 = 210$ systems. However, there are two exceptions to consider, which will have more than one solution. The first is of the form $x + y + z = 0$ $x + y = 0$ $z = 0$ ; the second is of the form $x = 0$ $y = 0$ $x + y = 0$ . In both cases, there are $3$ ways to choose the variables in the equations, and then $6$ ways to arrange them, giving $2 \cdot 3 \cdot 6 = 36$ exceptions. Subtracting this gives $210 - 36 = 174$ systems with only one solution, and the answer is then $512 - 174 = \boxed{338}$
| 338
|
1,857
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18
| 4
|
Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$
|
Denote vector $\overrightarrow{i} = \left( i_1, i_2, i_3 \right)^T$ for $i \in \left\{ a, b, c \right\}$ .
Thus, we need to count how many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly dependent.
We do complementary counting.
First, the total number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $\left( 2^3 \right)^3 = 512$
Second, we count how many many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly independent.
To meet this condition, no vector can be a zero vector $\overrightarrow{0} = \left( 0, 0, 0 \right)^T$
Next, we do the casework analysis.
Case $1^c$ : Three vectors are all on axes.
In this case, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $3!$
Case $2^c$ : Two vectors are on axes and the third vector is not.
We construct such an instance in the following steps.
Step 1: We determine which two vectors lie on axes.
The number of ways is $3$
Step 2: For two vectors selected in Step 1, we determine which two axes they lie on.
The number of ways is $3 \cdot 2$
Step 3: For the third unselected vector, we determine its value.
To make three vectors linear independent, the third vector cannot be on the plane formed by the first two vectors.
So the number of ways is $3$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 2 \cdot 3$
Case $3^c$ : One vector is on an axis and the other two are not.
We construct such an instance in the following steps.
Step 1: We determine which vector lies on an axis.
The number of ways is $3$
Step 2: For the selected vector, we determine which axis it lies on.
The number of ways is $3$
Step 3: We determine the values of the two unselected vectors.
First, to be linearly independent, these two vectors are distinct.
Second, to be linearly independent, we cannot have one vector $(1,1,1)$ and another one that is a diagonal vector on the plane that is perpendicular to the first selected vector.
Thus, the number or ways in this step is $4 \cdot 3-2 = 10$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 10$
Case $(4.4)^c$ : No vector is on any axis.
In this case, any three distinct vectors are linearly independent.
So the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $4 \cdot 3 \cdot 2$
Putting all cases together, the number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ that are linearly independent is \[ 8^3 - 3! - 3 \cdot 3 \cdot 2 \cdot 3 - 3 \cdot 3 \cdot 10 - 4 \cdot 3 \cdot 2 = \boxed{338}. \] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| 338
|
1,858
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_19
| 1
|
Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.) [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$
|
There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$
| 22
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1,859
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 1
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
Without loss of generality, we assume the length of each side of $ABCD$ is $2$ .
Because $E$ is the midpoint of $CD$ $CE = 1$
Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$
In $\triangle BCE$ , following from the law of sines, \[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]
We have $\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$
Hence, \[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]
By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$
Because $AF \perp BF$ \begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}
In $\triangle BFC$ , following from the law of sines, \[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]
Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$ , the equation above can be converted as \[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]
Therefore, \begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}
Therefore, $\angle BFC = \boxed{113}$
| 113
|
1,860
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 2
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$
Because $\overline{AB} \parallel \overline{ED}$ , we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$ , so $\triangle ABG \sim \triangle DEG$ by AA.
Because $ABCD$ is a rhombus, $AB = CD = 2DE$ , so $AG = 2GD$ , meaning that $D$ is a midpoint of segment $\overline{AG}$
Now, $\overline{AF} \perp \overline{BE}$ , so $\triangle GFA$ is right and median $FD = AD$
So now, because $ABCD$ is a rhombus, $FD = AD = CD$ . This means that there exists a circle from $D$ with radius $AD$ that passes through $F$ $A$ , and $C$
AG is a diameter of this circle because $\angle AFG=90^\circ$ . This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$ , so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$ , which means that $\angle BFC = \boxed{113}$
| 113
|
1,861
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 3
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
Let $\overline{AC}$ meet $\overline{BD}$ at $O$ , then $AOFB$ is cyclic and $\angle FBO = \angle FAO$ . Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$ , so $\frac{AF}{BO} = \frac{AC}{BE}$ , thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$ , then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$ , and $\angle BFC = \boxed{113}$
| 113
|
1,862
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 4
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$ . A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$ , meaning that $\angle BFC$ increases with $\angle D$ .
Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{113}$
| 113
|
1,863
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 5
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
This solution refers to the Diagram section.
We extend $AD$ and $BE$ to point $G$ , as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); [/asy] We know that $AB=AD=2$ and $CE=DE=1$
By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ . This implies that $2AD=AG$ and $AD \cong DG$ , so $AG=2AD=2\cdot2=4$ . That is, $D$ is the midpoint of $AG$
Now, let's redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$ , which is on the circle, as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NE,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); draw(Circle(D,6),dashed); [/asy] Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$
Let's call $\angle CFE = \theta$
Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$ , So $\angle CDG = 2\angle CFE$
Let $\angle CDG = 2\theta$ . Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that \begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67. \end{align*} Since $\angle BFC=180-\theta$ , we have $\angle BFC=180-67=\boxed{113}$
| 113
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1,864
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20
| 6
|
Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
|
If angle $ADC$ was a right angle, it would be much easier. Thus, first pretend that $ADC$ is a right angle. $ABCD$ is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line $AE$ , which is $\sqrt{5}/2$ . We want the measure of angle $BFC$ , so to work closer to it, we should try finding the length of line $BF$ . Angle $FAB$ and angle $ABF$ are complementary. Angle $ABF$ and angle $FBC$ are also complementary. Thus, $\sin FAB=\cos ABF=\sin FBC$ $\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}$ . Since $\sin FAB=1\sqrt{5}$ ,and $AB=1$ $FB=\sin FAB$ . It follows now that $FE=3*\sqrt{5}/10$
Now, zoom in on triangle $BEC$ . To use the Law of Cosines on triangle $FBC$ , we need the length of $FC$ . Use the Law of Cosines on triangle $EFC$ . Cos $E=1/\sqrt{5}$ . Thus, after using the Law of Cosines, $FC=\sqrt{2/5}$
Since we now have SSS on $BEC$ , we can get use the Law of Cosines. $\cos BFC=1/-\sqrt{2}$ $\arccos 1/-\sqrt{2}$ is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. $180-45=135$ . Angle $BFC$ is $135^\circ$
Realize that, around point F, there will always be 3 right angles, regardless of what angle $ADC$ is. There are only two angles that change when $ADC$ changes. Break up angle $BFC$ into angle $BFB'$ , which is always 90 degrees, and angle $B'FC$ , which we have discovered to to be half of $ADC$ . Thus, when angle $ADC$ is 46 degrees, then $B'FC$ will be 23. $23+90=113$ . Angle $BFC$ is $\boxed{113}$ degrees.
| 113
|
1,865
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 1
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.
Throughout this solution, we will express all polynomials in base $x$ . E.g. $x^2 + x + 1 = 111_{x}$
We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$ $b = 12_{x} = x + 2$ , and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$ $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$ ‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{23}$
| 23
|
1,866
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 2
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x+1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{23}$
| 23
|
1,867
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 3
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$ ,
then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$
Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$
We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$ , and $Q_2(x) = x + b$ , since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.
Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$
The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$
So $Q_1(x) - Q_2(x) = -1$
Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$
Solving this equation, we get $Q_1(x) = x + 1$
Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$
Verify this works with $P(x) = (x^2+1)Q_2(x) + 2x + 1$
Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{23}$
| 23
|
1,868
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 4
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let $P(x)=(x^2+x+1)(ax+b)+(x+2)$ and $P(x)=(x^2+1)(ax+c)+(2x+1)$ . The quotients have the same $x$ coefficient, since $P(x)$ must have the same $x^3$ coefficient in both cases. Expanding, we get \[P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)\] and \[P(x)=ax^3+cx^2+(a+2)x+(c+1).\]
Equating coefficients, we get $b+2=c+1$ $a+b+1=a+2$ , and $a+b=c$ . From the second equation, we get $b=1$ , then substituting into the first, $c=2$ . Finally, from $a+b=c$ , we have $a=1$ . Now, $P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3$ and our answer is \[1^2+2^2+3^2+3^2=\boxed{23}.\]
| 23
|
1,869
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 5
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x)$ are linear.
Let $Q(x)=x-q$ and $R(x)=x-r.$ We know there will be values for $q$ and $r$ that make the below equation hold, so we can assume that $P(x)$ has a leading coefficient of $1$
Substituting these values in, and setting $\textbf{(1)}$ and $\textbf{(2)}$ equal to each other, \[(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.\] We plug in $x=0$ , yielding $r+1=q.$ Substituting this value into the above equation, \[(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.\] Letting $x=1,$ we conclude that $r=-2,$ so $R(x)=x+2.$ Therefore, \[P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.\] The requested sum is \[1^2+2^2+3^2+3^2=\boxed{23}\]
| 23
|
1,870
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21
| 6
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
By remainder theorem, the polynomial can be written as follows.
\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both $Q_{1}(x)$ and $Q_{2}(x)$ are linear (the coefficient of x should be same).
Then we can write $P(x)$ as a cubic polynomial.
\[P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1\] Substituting $x=0,1,-1$ to determine the value of $a$ and $b$
We have: \[P(0) = b+2 = c+1\] \[P(1) = 3a+3b+3 = 2a+2c+3\] \[P(-1) = -a+b+1 = -2a+2c-1\]
We can solve the simultaneous equations: $a=1,b=1,c=2$
Hence, $P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3$ . The answer is $1^2+2^2+3^2+3^2=\boxed{23}$
| 23
|
1,871
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_22
| 1
|
Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$
|
We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ .
We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ .
We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$
We denote by $C_0$ a circle that is tangent to $C_1$ $C_2$ and $C_3$ .
We denote by $\left( u, v \right)$ the coordinates of circle $C_0$ , and $r$ the radius of this circle.
From the graphs of circles $C_1$ $C_2$ $C_3$ , we observe that if $C_0$ is tangent to all of them, then $C_0$ must be internally tangent to $C_2$ .
We have \[ u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1) \]
We do the following casework analysis in terms of the whether $C_0$ is externally tangent to $C_1$ and $C_3$
Case 1: $C_0$ is externally tangent to $C_1$ and $C_3$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 2: $C_1$ is internally tangent to $C_0$ and $C_3$ is externally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 3: $C_1$ is externally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 4: $C_1$ is internally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Because the graph is symmetric with the $x$ -axis, and for each case above, the solution of $v$ is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the $x$ -axis.
Therefore, the sum of the areas of all the circles in $S$ is $2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{136}$
| 136
|
1,872
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24
| 1
|
Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$
|
We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$
Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{50} ~MRENTHUSIASM
| 50
|
1,873
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24
| 2
|
Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$
|
Denote $f(900)-f(600) = a$ .
Because $f(300) = f(900)$ $f(300) - f(600) = a$
Following from the Lipschitz condition given in this problem, $|a| \leq 150$ and \[ f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\} \] and \[ f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} . \] Thus, \begin{align*} f(800) - f(400) & \leq \min \left\{ a + 50 , 100 \right\} - \max \left\{ a - 50 , -100 \right\} \\ & = 100 + \min \left\{ a, 50 \right\} - \max \left\{ a , - 50 \right\} \\ & = 100 + \left\{ \begin{array}{ll} a + 50 & \mbox{ if } a \leq -50 \\ 0 & \mbox{ if } -50 < a < 50 \\ -a + 50 & \mbox{ if } a \geq 50 \end{array} \right. . \end{align*} Thus, $f(800) - f(400)$ is maximized at $a = 0$ $f(800)-f(600) = 50$ $f(400)-f(600)=-50$ , with the maximal value 100.
By symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$ $f(800)-f(600) = -50$ $f(400)-f(600)=50$ , with the minimal value $-100$
Following from the Lipschitz condition, \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*} We have already construct instances in which the second inequality above is augmented to an equality.
Now, we construct an instance in which the first inequality above is augmented to an equality.
Consider the following piecewise-linear function: \[ f(x) = \left\{ \begin{array}{ll} \frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\ -\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\ \frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\ -\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800 \end{array} \right.. \] Therefore, the maximum value of $f(f(800)) - f(f(400))$ is $\boxed{50}$
| 50
|
1,874
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24
| 3
|
Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$
|
Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$ . This means that when we take any two points on $f$ , the absolute value of the slope between the two points is at most $\frac{1}{2}$
Let $f(300) = f(900) = c$ , and since we want to find the maximum value of $|f(800) - f(400)|$ , we can take the most extreme case and draw a line with slope $-\frac{1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\frac{1}{2}$ up from $f(800)$ to $f(900)$ . Then $f(400) = c - 50$ and $f(800) = c + 50$ , so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$ , and this is attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value less than $\frac{1}{2}$
Therefore, $|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{50}$
| 50
|
1,875
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24
| 4
|
Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$
|
Consider $g(x) = f(x)-f(300)$ . Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$ . We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$ , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50 )| \leq \frac{1}{2}(100) = \boxed{50}$ as the answer.
| 50
|
1,876
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25
| 1
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{6}.\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| 6
|
1,877
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25
| 2
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$ \[0 \equiv 2^{2019}k + 1 \pmod{7}.\] Using Fermat's Little Theorem, \[2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.\] Thus, \[0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.\] Therefore, \[7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.\]
We may repeat this same calculation with $S_{2023}$ to yield \[S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.\] Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits.
This gives rise to a simple inequality, \[0 \leqslant S_n \leqslant 2^n.\] Since the maximum possible number that can be generated with $n$ bits is \[\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.\] Looking at our calculations for $S_{2019}$ and $S_{2023}$ , we see that the only valid integers that satisfy that constraint are $j = h = 0$ \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{6}.\] ~zoomanTV
| 6
|
1,878
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25
| 4
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0. A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{6}.$
| 6
|
1,879
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1
| 1
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{49}.\] ~MRENTHUSIASM
| 49
|
1,880
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1
| 2
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{49}.\]
| 49
|
1,881
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1
| 3
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator is actually bigger than it should be, it should be the smaller answer, or $\boxed{49}$
| 49
|
1,882
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_2
| 1
|
Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$
|
We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\] Therefore, the answer is $\boxed{20}.$
| 20
|
1,883
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3
| 1
|
What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$
|
The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$
Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\] Approximating with $\pi\approx3.14,$ we have $12<4\pi<13,$ or $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor.$ We simplify to get \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$
| 6
|
1,884
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3
| 2
|
What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$
|
As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
By an underestimation $\pi\approx3,$ we have $4\pi>12,$ or $\frac{81}{4\pi}<6\frac34.$
By an overestimation $\pi\approx\frac{22}{7},$ we have $4\pi<\frac{88}{7},$ or $\frac{81}{4\pi}>6\frac{39}{88}.$
Together, we get \[6 < 6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34 < 7,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$
| 6
|
1,885
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3
| 3
|
What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$
|
As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
Approximating with $\pi\approx3,$ we have $\frac{81}{4\pi}\approx6\frac34.$ Since $\pi$ is about $5\%$ greater than $3,$ it is safe to claim that $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$
| 6
|
1,886
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5
| 1
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
First, modulo $2$ or $5$ $\underline{20210A} \equiv A$ .
Hence, $A \neq 0, 2, 4, 5, 6, 8$
Second modulo $3$ $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$ .
Hence, $A \neq 1, 4, 7$
Third, modulo $11$ $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$ .
Hence, $A \neq 3$
Therefore, the answer is $\boxed{9}$
| 9
|
1,887
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5
| 2
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\textbf{(C)}$
If the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ . Thus we can eliminate options $\textbf{(A)}$ and $\textbf{(D)}$
So the correct option is either $\textbf{(B)}$ or $\textbf{(E)}$ . Let's try dividing the number with some integers.
$20210A/7 = 2887x$ , where $x$ is $1A/7$ . Since $13$ and $19$ are both indivisible by $7$ , this does not help us narrow the choices down.
$20210A/11 = 1837x$ , where $x$ is $3A/11$ . Since $33/11 = 3$ , option $\textbf{(B)}$ would make $20210A$ divisible by $11$ . Thus, by elimination, the correct choice must be option $\boxed{9}$
| 9
|
1,888
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5
| 3
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
$202100 \implies$ divisible by $2$
$202101 \implies$ divisible by $3$
$202102 \implies$ divisible by $2$
$202103 \implies$ divisible by $11$
$202104 \implies$ divisible by $2$
$202105 \implies$ divisible by $5$
$202106 \implies$ divisible by $2$
$202107 \implies$ divisible by $3$
$202108 \implies$ divisible by $2$
This leaves only $A=\boxed{9}$
| 9
|
1,889
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_6
| 1
|
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
|
There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride.
| 8
|
1,890
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_6
| 2
|
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
|
There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$
| 8
|
1,891
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_7
| 1
|
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
|
By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{170}$ degrees.
| 170
|
1,892
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_7
| 2
|
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
|
We can extend $\overline{AD}$ to $G$ , making $\angle CDG$ a right angle. It follows that $\angle GDE$ is $110^\circ - 90^\circ = 20^\circ$ , as shown below. [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); pair G = (15,10); label("$G$", G, E); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F^^G); draw(A--B--C--D--G--cycle); draw(D--EE--F--cycle); [/asy] Since $\angle DFE = \angle DEF$ , we see that $\angle DFE = \angle DEF = \frac{20}{2} = 10^\circ$ . Thus, $\angle AFE = 180^\circ - 10^\circ = \boxed{170}$ degrees.
| 170
|
1,893
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11
| 1
|
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$
|
Let $x$ be the length of the ship.
Then, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps.
Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps.
Since the ship and Emily have the same ratio of absolute speeds in either direction, $\frac{210}{210-x} = \frac{42}{x-42}$ . Dividing both sides by $42$ and cross multiplying, we get $5(x-42) = 210-x$ , so $6x = 420$ , and $x = \boxed{70}$
| 70
|
1,894
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11
| 2
|
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$
|
Let the speed at which Emily walks be $42$ steps per hour. Let the speed at which the ship is moving be $s$ . Walking in the direction of the ship, it takes her $210$ steps, or $\frac {210}{42} = 5$ hours, to travel. We can create an equation: \[d = 5(42-s),\] where $d$ is the length of the ship. Walking in the opposite direction of the ship, it takes her $42$ steps, or $42/42 = 1$ hour. We can create a similar equation: \[d = 1(42+s).\] Now we have two variables and two equations. We can equate the expressions for $d$ and solve for $s$ \begin{align*} 210-5s &= 42 + s \\ s &= 28. \\ \end{align*} Therefore, we have $d = 42 + s = \boxed{70}$
| 70
|
1,895
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11
| 3
|
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$
|
Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.
Let $L$ be the length of the ship, $E$ be Emily's step length, and $S$ be the ship's step length. We wish to find $\frac LE.$
When Emily walks from the back of the ship to the front, she walks a distance of $210E$ and the front of the ship moves a distance of $210S.$ We have $210E=L+210S$ for this scenario, which rearranges to \[210E-210S=L. \hspace{15mm}(1)\] When Emily walks in the opposite direction, she walks a distance of $42E$ and the back of the ship moves a distance of $42S.$ We have $42E=L-42S$ for this scenario, which rearranges to \[42E+42S=L. \hspace{19.125mm}(2)\] We multiply $(2)$ by $5$ and then add $(1)$ to get $420E=6L,$ from which $\frac LE = \boxed{70}.$
| 70
|
1,896
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11
| 4
|
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$
|
Every time Emily takes a step, the boat also "takes a step". Call the length of the boats step $s$ . Call the length of the boat $x$
When Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of $s$ . This means that she travels a total distance of $x + 210 s$ to reach the other end of the boat.
When Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by $s$ (since the boat is coming towards her and moves a distance of $s$ ). This means that she travels a total distance of $x - 42 s$ to reach the other end of the boat.
Taking Emily's step as a unit of distance, we now have two equations \begin{align*} 210 &= x + 210 s, \\ 42 &= x - 42s. \end{align*} Solving for $x$ you get $\boxed{70}$
| 70
|
1,897
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11
| 5
|
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$
|
Call the speed of the boat $v_s$ and the speed of Emily $v_e$
Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is $v_e-v_s$
Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is $v_e+v_s$
Since Emily takes $210$ steps to walk along with the boat and $42$ steps to walk opposite the boat, that means it takes her $5$ times longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$
This means that $5(v_e-v_s)=v_e+v_s$ , so $v_s = \frac{2v_e}{3}$
As Emily takes $210$ steps to walk the length of the boat at a speed of $v_e- \frac{2v_e}{3}=\frac{v_e}{3}$ , she must take $\frac13$ of the time to walk the length of the boat at a speed of $v_e$ , so our answer is $\frac{210}{3} = \boxed{70}$
| 70
|
1,898
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_14
| 1
|
How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$
|
The second equation is $(|x|+|y| - 4)^2 = 1$ . We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: \[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\] We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane: [asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy] We see from the graph that there are $5$ intersections, so the answer is $\boxed{5}$
| 5
|
1,899
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15
| 1
|
Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$
|
Let $\odot O_1$ be the circle with radius $5\sqrt2$ that is tangent to $\overleftrightarrow{AB}$ at $B$ and to $\overleftrightarrow{AC}$ at $C.$ Note that $\angle ABO_1 = \angle ACO_1 = 90^\circ.$ Since the opposite angles of quadrilateral $ABO_1C$ are supplementary, quadrilateral $ABO_1C$ is cyclic.
Let $\odot O_2$ be the circumcircle of quadrilateral $ABO_1C.$ It follows that $\odot O_2$ is also the circumcircle of $\triangle ABC,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O1, O2; A = (0,2sqrt(26)); O1 = (0,0); B = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[0]; C = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[1]; O2 = midpoint(A--O1); fill(A--B--C--cycle, yellow); dot("$A$",A,1.5*N,linewidth(4)); dot("$B$",B,1.5*W,linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$O_1$",O1,1.5*S,linewidth(4)); dot("$O_2$",O2,1.5*N,linewidth(4)); label("$3\sqrt6$",midpoint(A--B),scale(0.5)*rotate(90)*dir(midpoint(A--B)--A),red+fontsize(10)); label("$3\sqrt6$",midpoint(A--C),scale(0.5)*rotate(90)*dir(midpoint(A--C)--C),red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--B),0.5*SW,red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--C),0.5*SE,red+fontsize(10)); markscalefactor=0.05; draw(rightanglemark(A,B,O1)^^rightanglemark(A,C,O1),red); draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); [/asy] By the Inscribed Angle Theorem, we conclude that $\overline{AO_1}$ is the diameter of $\odot O_2.$ By the Pythagorean Theorem on right $\triangle ABO_1,$ we have \[AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.\] Therefore, the area of $\odot O_2$ is $\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{26}.$
| 26
|
1,900
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15
| 2
|
Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$
|
[asy] import olympiad; unitsize(50); pair A,B,C,D,E,I,F,G,O; A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,D,E); draw(A--B--C--cycle); dot(O); dot(I); dot(F); dot(G); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,D,E)); draw(I--B); draw(I--C); draw(I--A); draw(rightanglemark(A,C,I)); draw(rightanglemark(A,B,I)); draw(O--F); draw(O--G); draw(rightanglemark(A,F,O)); draw(rightanglemark(A,G,O)); label("$O$",O,W); label("$A$",A,S); label("$B$",B,N); label("$C$",C,W); label("$D$",F,S); label("$E$",G,W); label("$3\sqrt{6}$",(1.5,1.5),S); label("$3\sqrt{6}$",(-1.5,1.5),S); label("$5\sqrt{2}$",(1,3.625),N); label("$5\sqrt{2}$",(-1,3.625),N); label("$I$",I,N); label("$r$",(-0.25,1.5),E); label("$r$",(0.5,2.125),S); add(pathticks(A--F,1,0.5,0,2)); add(pathticks(F--B,1,0.5,0,2)); add(pathticks(A--G,1,0.5,0,2)); add(pathticks(G--C,1,0.5,0,2)); [/asy] Because circle $I$ is tangent to $\overline{AB}$ at $B, \angle{ABI} \cong 90^{\circ}$ . Because $O$ is the circumcenter of $\bigtriangleup ABC, \overline{OD}$ is the perpendicular bisector of $\overline{AB}$ , and $\angle{BAI} \cong \angle{DAO}$ , so therefore $\bigtriangleup ADO \sim \bigtriangleup ABI$ by AA similarity. Then we have $\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}$ . We also know that $\overline{AD} = \frac{3\sqrt{6}}{2}$ because of the perpendicular bisector, so the hypotenuse of $\bigtriangleup ADO$ is \[\sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2+\left(\frac{3\sqrt{6}}{2}\right)^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}.\] This is the radius of the circumcircle of $\bigtriangleup ABC$ , so the area of this circle is $\boxed{26}$
| 26
|
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