problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
There are $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to ... | B | 420 |
40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | We can choose a different frame to solve this problem. Our tile combination can be written as $Y, Y, Y, G, G, B, P.$ We can focus on $Y, Y, Y, G, G$ first, which gives us $\binom{5}{3}=10.$ Now we can insert our brown tile into this which only has $6$ choices(like $Y,B, Y, Y, G, G$ and $Y, Y, Y, G, G, B$ etc.), then in... | B | 420 |
40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | Let $B$ be brown, $P$ be purple, $G$ be green, and $Y$ be yellow. Then, we are just ordering $Y$ $Y$ $Y$ $G$ $G$ $B$ , and $P$ . Hence, $\frac{7!}{3! \cdot 2!} = \boxed{420}$ | B | 420 |
39a604ab82f19405c40da69abfbddc54 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_6 | Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$ -hour period?
$\textbf{(A)}\ 50 \qq... | In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.
So we raise $9$ to the next largest value, $10$ , ... | B | 55 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | Any even multiple of $3$ is a multiple of $6$ , so we need to find multiples of $6$ that are perfect squares and less than $2020$ . Any solution that we want will be in the form $(6n)^2$ , where $n$ is a positive integer. The smallest possible value is at $n=1$ , and the largest is at $n=7$ (where the expression equals... | A | 7 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$ , where $3^2$ is the multiple or $3$ and $2^2$ makes it even. Simplifying we have $36^2 \cdot x^2$ . We can divide $2020$ by $36$ (floor) and get $56$ see the result. We can then see that there are $7$ different values for $x$ . It can't be l... | A | 7 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$ $12^2$ $18^2$ $24^2$ $30^2$ $36^2$ , and $42^2$ . Since $48^2=2196 > 2020$ , there are $\boxed{7}$ valid answers.
~airbus-a321, November 2023 | A | 7 |
e0fb72b6e1b8e3785100a2de5fa7715f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 | Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:
[asy] /* Made by MRENTHUSIASM */ size(250); pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2... | D | 8 |
e0fb72b6e1b8e3785100a2de5fa7715f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 | Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
Without the loss of generality, let $P=(-4,0)$ and $Q=(4,0).$ We conclude that the $y$ -coordinate of $R$ must be $\pm3.$
We apply casework to the right angle of $\triangle PQR:$
Together, ther... | D | 8 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging ... | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are ... | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power... | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $... | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This g... | D | 4 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | We have
\[\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]
Now we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\underbrace{00\dots0}_{26 \text... | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$ . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$
$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fa... | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the... | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | We see that $\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}$ . We see that this has $27-1=26$ zeros after the decimal point before coming to $9$
Therefore, the answer is $\boxed{26}$ | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | \begin{align*}|\lceil \log \dfrac{1}{20^{20}} \rceil| &= |\lceil \log 20^{-20} \rceil| \\ &= |\lceil -20 \log(20) \rceil| \\ &= |\lceil -20(\log 10 + \log 2) \rceil| \\ &= |\lceil -20(1 + 0.301) \rceil| \\ &= |\lceil -26.02 \rceil| \\ &= |-26| \\ &= \boxed{26} | D | 26 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:
Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: \[(12\underline{425}3415251).\] Therefore, ... | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | After erasing every third digit, the list becomes $1245235134\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\ldots$ repeated. Since this list repeats every $12$ d... | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | Note that cycles exist initially and after each round of erasing.
We will consider one cycle after all three rounds of erasing. Suppose this cycle has length $L$ before any round of erasing. It follows that:
The least such positive integer $L$ is $\operatorname{lcm}(5,3,2,5)=30.$ So, there is a repeating pattern for ev... | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | As the LCM of $3$ $4$ , and $5$ is $60$ , let us look at a $60$ -digit block of original numbers (many will be erased by Steve). After he erases every third number $\left(\dfrac{1}{3}\right)$ , then every fourth number of what remains $\left(\dfrac{1}{4}\right)$ , then every fifth number of what remains $\left(\dfrac{1... | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | Lemma: Given a sequence $a_1, a_2, a_3, \cdots$ , and an positive integer $k>2$ . If we erase every $k$ th item in this sequence, and we name $b_1, b_2, b_3, \cdots$ as the remaining sequence.
Then we have \[b_{(k-1)m+1}=a_{km+1}, b_{(k-1)m+2}=a_{km+2}, \cdots, b_{(k-1)m+k-1}=a_{km+k-1}.\]
Proof: For $a_{km+j}$ with s... | D | 11 |
667c9a200bc443791b4e290668337093 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know... | Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: $0$ diameters
There are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1... | C | 13 |
667c9a200bc443791b4e290668337093 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know... | If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply ... | C | 13 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.
Now, suppose that we have this deck again, with only one black card. Each time we pic... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Let $R$ denote the action where George selects a red ball and $B$ denote the action where he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$
There are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\binom{4}{2}=6$ ). H... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | We know that we need to find the probability of adding 2 red and 2 blue balls in some order.
There are 6 ways to do this, since there are $\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.
We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted b... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red b... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\t... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Here $X$ stands for R or B, and $Y$ for the remaining color.
After 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \[\frac{2}{3} \... | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | We can use dynamic programming to solve this problem.
We let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.
Notice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$... | B | 15 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | $158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$
We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cd... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
And after simplifying, we get $26\cdot17\cdot7\cdo... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43$
Since this number is divisible by $4$ but not $8$ , the last $2$ digits must be divisible by $4$ but the last $3$ digits c... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | It is not hard to check that $13$ divides the number, \[\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.\] As $10^3\equiv-1\pmod{13}$ , using $\pmod{13}$ we have $13|\overl... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | As mentioned above, \[\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.\] We can divide both sides of $1... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | We note that: \[\frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43).\] Let $K=(13)(17)(7)(47)(46)(5)(22)(43)$ . This will help us find the last two digits modulo $4$ and modulo $25$ .
It is obvious that $K \equiv 0 \pmod{4}$ . Also (although this not so obvi... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | As in Solution 2, we see that
\[\binom{52}{10}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43,\]
which contains no factors of $3.$ Therefore, the sum of the digits must not be a multiple of $3.$ This sum is
\[1+5+8+A+0+0+A+4+A+A+0=18+4A.\]
It follows that $4A$ cannot be a multiple of $3,$ ruling out choices $(B)$ an... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | The total number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Using divisibility rules, we have that A is not a multiple of $3$ . Then, divide this equation by 10. This implies that... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | As stated in previous solutions, the number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Canceling out common factors $(\dfrac{52}{2} = 26 \text{, } \dfrac{51}{3} = 17 \text{, } 5 \... | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | Compute $\frac{52!}{10!42!} = 15820024220.$ Therefore our answer is $\boxed{2}.$ | A | 2 |
80d780837fb2c8c088bc8f294741bac0 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_20 | Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$ , together with its interior. For real $r\geq0$ , let $S(r)$ be the set of points in $3$ -dimensional space that lie within a distance $r$ of some point in $B$ . The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$ , where $a,... | Split $S(r)$ into 4 regions:
1. The rectangular prism itself
2. The extensions of the faces of $B$
3. The quarter cylinders at each edge of $B$
4. The one-eighth spheres at each corner of $B$
Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$ , so $d=12$
Region 2: This volume is equal to the surface area of $B$ ti... | B | 19 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Completing the square, then difference of squares:
\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ &= (2^{101} + 1)^2 - 2^{102} + 201\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. \end{align*}
Thus, we see that the remainder is surely $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Let $x=2^{50}$ . We are now looking for the remainder of $\frac{4x^4+202}{2x^2+2x+1}$
We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity , which states that \[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]
Let's use the identity, with $a=1$ and $b=x$ , so we have... | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We let \[x = 2^{50}\] and \[2^{202} + 202 = 4x^{4} + 202\] .
Next we write \[2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1\] .
We know that \[4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)\] by the Sophie Germain identity so to find \[4x^{4} + 202,\] we find that \[4x^{4} + 202 = 4x^{4} + 201 + 1\] which shows that the remai... | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We let $x=2^{50.5}$ . That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Let $n=2^{101}+2^{51}+1$ . Then, mod $n$
$2^{202}+202 \equiv (-2^{51}-1)^2 + 202$
$\equiv 2^{102}+2^{52}+203$
$= 2(n-1)+203 \equiv 201 \pmod{n}$
Thus, the remainder is $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
\[\frac{2^{202}+202}{2^{101}+2^{51}+1}\] \[= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} ... | D | 201 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = ... | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | This is my first solution here, so please forgive me for any errors.
We are given that \[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\]
$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ . As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+5... | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separat... | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lf... | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le 70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which ... | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not squ... | C | 6 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmet... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization nee... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ . An arrangement of factors can be created by placing "dividers" to group primes. For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ . Because there are $6$ ways to order the prime factors, and... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}... | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\c... | null | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero... | A | 112 |
91e01dfe831ce598f96a687a425bc3b7 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_1 | What is the value of \[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | $2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$
$= 1+1 = \boxed{2}$ | C | 2 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,
\[A-1 = 5(B-1)\] and \[A = B^2.\]
Substituting the second equation into the first gives us
\[B^2-1 = 5(B-1).\]
By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$
The answer is $16-4 = 12 \implies \boxed{12}$ | D | 12 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | Simple guess and check works. Start with all the square numbers - $1$ $4$ $9$ $16$ $25$ $36$ , etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$ , then Bonita is $3$ , so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$ , then Bonita is $4$... | D | 12 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $12 \rightarrow \boxed{12}.$ | D | 12 |
fc30ba5fbc0078b42ae28e42306bab6c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_4 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) }... | We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$... | B | 76 |
c6ba51c90cefa9e1c985fd8b02333488 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_5 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\b... | D | 90 |
c6ba51c90cefa9e1c985fd8b02333488 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_5 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$ | D | 90 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer... | C | 3 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is $\boxed{3}$ | C | 3 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is $\b... | C | 3 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ .
Now we find the i... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 ... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\sqrt2$ and equal length $2\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\frac{b\sqrt{4l^2-b^2}}{4}$ . ... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | The area of a triangle formed by three lines, \[a_1x + a_2y + a_3 = 0\] \[b_1x + b_2y + b_3 = 0\] \[c_1x + c_2y + c_3 = 0\] is the absolute value of \[\frac12 \cdot \frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \cdot \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}^2\] Plugging in the t... | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$ . Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqr... | C | 6 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ , and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$ , which means they would be able to cancel with the f... | B | 996 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$ . Choices $A$ $C$ , and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which... | B | 996 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$ . The quotient of the two is -
$\frac{(2)(n!)}{(n)(n+1)}$
which simplifies to $\frac{(2)((n-1)!)}{n+1}$ . Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some num... | B | 996 |
37a52accb87db88639f07386c5011bd6 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_10 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad... | The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line. As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \boxed{26}$ squares. | C | 26 |
37a52accb87db88639f07386c5011bd6 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_10 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad... | We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ ... | C | 26 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Prime factorizing $201^9$ , we get $3^9\cdot67^9$ .
A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$ . This yields $5\cdot5 = 25$ perfect squares.
Perfect cubes must have multiples of $3$ for each of t... | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Observe that $201 = 67 \cdot 3$ . Now divide into cases:
Case 1 : The factor is $3^n$ . Then we can have $n = 2$ $3$ $4$ $6$ $8$ , or $9$
Case 2 : The factor is $67^n$ . This is the same as Case 1.
Case 3 : The factor is some combination of $3$ s and $67$ s.
This would be easy if we could just have any combination, as ... | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | We first prime factorize $201^9 = 3^9 \cdot 67^9$ . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of $3$ in the exponent. The largest square for $3$ can be $3^8$ , so their must be $\dfrac {8}{2} = 4$ ways. The largest cube is $3^9$ , so there mu... | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Notice that $201=3 \cdot 67$ . We factorize $201^9$ to get $3^9 \cdot 67^9$ . We then list perfect squares and cubes. $3^2$ $3^4$ $3^6$ $3^8$ $3^3$ $3^6$ $3^9$ $67^2$ $67^4$ $67^6$ $67^8$ $67^3$ $67^6$ $67^9$ . Notice that the powers of $6$ overlap. We must not forget $1$ though. Of course, all of these factors already... | C | 37 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | [asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]
Drawing it out, we see $\angle BDC$ and $\angle BE... | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, \[\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.\]
So, $\triangle BEF \sim BDA$ ... | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | Through the property of angles formed by intersecting chords, we find that \[m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}\]
Through the Outside Angles Theorem, we find that \[m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}\]
Adding the two equations gives us \[m\angle BFC + m\angle CAB = m\overarc{BC}\implies ... | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | Notice that if $\angle BEC = 90^{\circ}$ , then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$ . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$ . Thus $\angle CBF = 70 - 20 = 50^{\circ}$ , and so $\angle BFC = 180 - 20 - 50 =... | D | 110 |
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