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f71a937f00c4ba287adc39caa511d37d
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_17
|
In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$
$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$
|
Let $AB$ , or the side of the octagon, be $x$ . Then, $BQ = \left(\frac{8-x}{2}\right)$ and $CQ = \left(\frac{6-x}{2}\right)$ . By the Pythagorean Theorem $BQ^2+CQ^2=x^2$ , or $\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2$ . Multiplying this out, we have $x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}$ . Simplifying, $-2x^2-28x+100=0$ . Dividing both sides by $-2$ gives $x^2+14x-50=0$ . Therefore, using the quadratic formula , we have $x=-7 \pm 3\sqrt{11}$ . Since lengths are always positive, then $x=-7+3\sqrt{11} \Rightarrow k+m+n=-7+3+11=\boxed{7}$
|
B
|
7
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.
Case $1$ :
Second Row: a b c
Third Row: b c a
Case $2$ :
Second Row: a c b
Third Row: c b a
Case $3$ :
Second Row: a b c
Third Row: c a b
Case $4$ :
Second Row: a c b
Third Row: b a c
For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has $2 \cdot 2 \cdot 2 = 8$ possibilities. Since there are four cases, when pair $a$ has someone in the leftmost seat of the second row, there are $32$ ways to rearrange it. However, someone from either pair $a$ $b$ , or $c$ could be sitting in the leftmost seat of the second row. So, we have to multiply it by $3$ to get our answer of $32 \cdot 3 = 96$ . So, the correct answer is $\boxed{96}$
|
D
|
96
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
Lets call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$ . We can split our problem into two cases:
There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.
Starting off with the first case, we see that there are $3!=6$ ways to arrange the A,B,C. Then, we have to choose which sibling sits. There are $2$ choices for each set of siblings meaning we have $2^3=8$ ways to arrange that. So, there are $48$ ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are $2$ ways to place the A sibling and each placement yields only $1$ possibility. So, our first case has $48\cdot2=96$ possibilities.
In our second case, there are $3$ ways to choose which set of siblings will be in the same row and $4$ ways to choose the person in between. So, there $3*4 = 12$ ways to arrange the first row. In the second row, however, we see that it is impossible to make everything work out. So, there are $0$ possibilities for this case.
Thus, there are $96+0 = \boxed{96}$ possibilities for this trip.
|
D
|
96
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
Call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it ( $\circ$ denotes an empty seat):
\[A_1 \circ \circ\] \[\circ \ \circ \ \circ\]
Then there are 4 choices for the second seat ( $B_1$ $B_2$ $C_1$ , or $C_2$ ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$
\[A_1 B_1 \circ \\\] \[\circ \ \circ \ \circ\]
The last seat in the first row cannot be $A_2$ because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be $C_1$ or $C_2$ . Let's say WLOG that it is $C_1$ . There are two ways to create a second row:
\[A_1 B_1 C_1 \\\] \[B_2 C_2 A_2\]
and
\[A_1 B_1 C_1 \\\] \[C_2 A_2 B_2\]
Therefore, there are $6 \cdot 4 \cdot 2 \cdot 2= \boxed{96}$ possible seating arrangements.
|
D
|
96
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
WLOG, define the three pairs of siblings to be: $A$ $B$ , and $C$ . Now, notice that you can only form a correct grouping either like this:
\[A B C\]
\[B C A\]
or this:
\[C B A\]
\[A C B\]
However, we need to consider the different orders. There are $3!$ ways to order each pair (eg. the same letters) and $2^3$ ways to order the people each of the three pairs. Now, we can just multiply everything together, yielding:
\[2\cdot3!\cdot2^3\]
Which is $\boxed{96}$
|
D
|
96
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
Let the families be $A$ $B$ $C$ . In any given possible arrangement, there are $3! = 6$ ways to arrange the families and $2 \cdot 2 \cdot 2 = 8$ ways to arrange the siblings. This means the answer has to be divisble by $6 \cdot 8 = 48$ . The only answer choice that satisfies this is $\boxed{96}$
|
D
|
96
|
20cefac0e0999db3223a72230f453eb7
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18
|
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$
|
If a pair of siblings are in the same row, the other $2$ pairs of siblings cannot fit in the remaining $4$ seats to meet the requirements. Therefore, the siblings must be in different rows.
Let the first pair of siblings be $a_1$ $a_2$ , the second pair of siblings be $b_1$ $b_2$ , and the third pair of siblings be $c_1$ $c_2$ . For each row there must be $1$ child from each family. WLOG, let $a_1$ $b_1$ $c_1$ be in the first row. There are $3!$ arrangements. $a_2$ $b_2$ $c_2$ are in the second row, and they cannot be in the same column as their sibling. Now the problem becomes a $3$ element Derangement problem.
An $n$ element derangement problem is to find the number of permutations of $n$ elements where each element has a specified location, and no element is in it's specified location. For example, there are $3$ elements $1$ $2$ $3$ $1$ cannot be in the first location, $2$ cannot be in the second location, and $3$ cannot be in the third location. The derangements are: $(2,3,1)$ $(3,1,2)$ $D_3=2$
Each pair of siblings can swap their position. There are $2^3$ swaps. The total arrangements are: $3! \cdot D_3 \cdot 2^3=6 \cdot 2 \cdot 8 =\boxed{96}$
|
D
|
96
|
ee3d21b54d77b312affa5f495e08ab9d
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19
|
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.
Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{11}.$
|
E
|
11
|
ee3d21b54d77b312affa5f495e08ab9d
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19
|
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solutions for $c-1$ . We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$ , so $c=37$ . Therefore, $j=38$ . Now, since $j-1=37$ , by similar logic, $37=(1+k)(a-1)$ , so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{11}$
|
E
|
11
|
ee3d21b54d77b312affa5f495e08ab9d
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19
|
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$ , so Chloe is $37$ , and Joey is $38$ . The common factor that will divide both of their ages is $37$ , so Joey will be $74$ . The answer is $7 + 4 = \boxed{11}$
|
E
|
11
|
ee3d21b54d77b312affa5f495e08ab9d
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_19
|
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$ , it will hit every natural number less than $C-Z$ , including every factor of $C-Z$ . For $C+n$ to be an integral multiple of $Z+n$ , the difference $C-Z$ must also be a multiple of $Z$ , which happens if $Z$ is a factor of $C-Z$ . Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$ . Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$ . Because $37$ is prime, the next time this happens is at $Z+n=37$ , when $J+n=74$ . The answer is $7+4=\boxed{11}$
|
E
|
11
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
For all integers $n \geq 7,$ note that \begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ &=-f(n-3)+2n-1 \\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ &=-f(n-4)+f(n-5)+n+2 \\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ &=f(n-6)+6. \end{align*} It follows that \begin{align*} f(2018)&=f(2012)+6 \\ &=f(2006)+12 \\ &=f(2000)+18 \\ & \ \vdots \\ &=f(2)+2016 \\ &=\boxed{2017} ~MRENTHUSIASM
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
For all integers $n\geq3,$ we rearrange the given equation: \[f(n)-f(n-1)+f(n-2)=n. \hspace{28.25mm}(1)\] For all integers $n\geq4,$ it follows that \[f(n-1)-f(n-2)+f(n-3)=n-1. \hspace{15mm}(2)\] For all integers $n\geq4,$ we add $(1)$ and $(2):$ \[f(n)+f(n-3)=2n-1. \hspace{38.625mm}(3)\] For all integers $n\geq7,$ it follows that \[f(n-3)+f(n-6)=2n-7. \hspace{32mm}(4)\] For all integers $n\geq7,$ we subtract $(4)$ from $(3):$ \[f(n)-f(n-6)=6. \hspace{47.5mm}(5)\] From $(5),$ we have the following system of $336$ equations: \begin{align*} f(2018)-f(2012)&=6, \\ f(2012)-f(2006)&=6, \\ f(2006)-f(2000)&=6, \\ & \ \vdots \\ f(8)-f(2)&=6. \end{align*} We add these equations up to get \[f(2018)-f(2)=6\cdot336=2016,\] from which $f(2018)=f(2)+2016=\boxed{2017}.$
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
Preamble: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$ th term of the sequence $A.$ This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.
To begin, we consider the sequence $B$ formed when we take the difference of consecutive terms between $A.$ Define $b_n = a_{n+1} - a_n.$ Notice that for $n \ge 4,$ we have
Notice that subtracting the second equation from the first, we see that $b_{n} = b_{n-1} - b_{n-2} + 1.$
If you didn’t notice that $B$ repeated directly in the solution above, you could also, possibly more naturally, take the finite differences of the sequence $b_n$ so that you could define $c_n = b_{n+1} - b_n.$ Using a similar method as above through reindexing and then subtracting, you could find that $c_n = c_{n-1} - c_{n-2}.$ The sum of any six consecutive terms of a sequence which satisfies such a recursion is $0,$ in which you have that $b_{n} = b_{n+6}.$ In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of $C$ to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence $C$ looks like \[\underbrace{2, 1, -1, -2, -1, 1,}_{\text{cycle period}} 2, 1, -1, -2, -1, 1, \ldots\] in which the same result follows.
Using the fact that $B$ repeats every six terms, this motivates us to look at the sequence $B$ more carefully. Doing so, we see that $B$ looks like \[\underbrace{2, 3, 2, 0, -1, 0,}_{\text{cycle period}} 2, 3, 2, 0, -1, 0, \ldots\] (If you tried pattern finding on sequence $B$ directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)
Now, there are two ways to finish.
Finish Method #1: Notice that any six consecutive terms of $B$ sum to $6,$ after which we see that $a_n = a_{n-6} + 6.$ Therefore, $a_{2018} = a_{2012} + 6 = \cdots = a_{2} + 2016 = \boxed{2017}.$
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
Start out by listing some terms of the sequence. \begin{align*} f(1)&=1 \\ f(2)&=1 \\ f(3)&=3 \\ f(4)&=6 \\ f(5)&=8 \\ f(6)&=8 \\ f(7)&=7 \\ f(8)&=7 \\ f(9)&=9 \\ f(10)&=12 \\ f(11)&=14 \\ f(12)&=14 \\ f(13)&=13 \\ f(14)&=13 \\ f(15)&=15 \\ & \ \vdots \end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be $+2$ $+3$ $+2$ $+0$ $-1$ $+0$ .
The largest odd multiple of $3$ smaller than $2018$ is $2013$ , so we have \begin{align*} f(2013)&=2013 \\ f(2014)&=2016 \\ f(2015)&=2018 \\ f(2016)&=2018 \\ f(2017)&=2017 \\ f(2018)&=\boxed{2017}
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
Writing out the first few values, we get \[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\ldots.\] We see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\pmod{6}$ and less than $2018$ is $2017,$ so we have $f(2017)=f(2018)=\boxed{2017}.$
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ f(n-1)&=f(n-2)-f(n-3)+n-1 \end{align*} Subtracting those two and rearranging gives \begin{align*} f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\ f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 \end{align*} Subtracting those two gives $f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.$
The characteristic polynomial is $x^4-3x^3+4x^2-3x+1=0.$
$x=1$ is a root, so using synthetic division results in $(x-1)(x^3-2x^2+2x-1)=0.$
$x=1$ is a root, so using synthetic division results in $(x-1)^2(x^2-x+1)=0.$
$x^2-x+1=0$ has roots $x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}.$
And \[f(n)=(An+D)\cdot1^n+B\cdot\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^n+C\cdot\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^n.\] Plugging in $n=1$ $n=2$ $n=3$ , and $n=4$ results in a system of $4$ linear equations $\newline$ Solving them gives $A=1, \ B=\frac{1}{2}-\frac{i\sqrt{3}}{2}, \ C=\frac{1}{2}+\frac{i\sqrt{3}}{2}, \ D=1.$ Note that you can guess $A=1$ by answer choices.
So plugging in $n=2018$ results in \begin{align*} 2018+1+\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^{2019}+\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{2019}&=2019+(\cos(-60^{\circ})+\sin(-60^{\circ}))^{2019})+(\cos(60^{\circ})+\sin(60^{\circ}))^{2019}) \\ &=2019+(\cos(-60^{\circ}\cdot2019)+\sin(-60^{\circ}\cdot2019))+(\cos(60^{\circ}\cdot2019)+sin(60^{\circ}\cdot2019)) \\ &=\boxed{2017} ~ryanbear
|
B
|
2017
|
3fed054befaf2a2b720d94b28fc07d89
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20
|
A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
|
We utilize patterns to solve this equation: \begin{align*} f(3)&=3, \\ f(4)&=6, \\ f(5)&=8, \\ f(6)&=8, \\ f(7)&=7, \\ f(8)&=8. \end{align*} We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern.
First, we need to know whether or not $2016$ is part of the skip or repeat. We notice that $f(6),f(12), \ldots,f(6n)$ all satisfy $6+6(n-1)=n,$ and we know that $2016$ satisfies this, leaving $n=336.$ Therefore, we know that $2016$ is part of the repeat section. But what number does it repeat?
We know that the repeat period is $2,$ and it follows that pattern of $1,1,8,8,7,7.$ Again, since $f(6) = f(5)$ and so on for the repeat section, $f(2016)=f(2015),$ so we don't need to worry about which one, since it repeats with period $2.$ We see that the repeat pattern of $f(6),f(12),\ldots,f(6n)$ follows $8,14,20,$ it is an arithmetic sequence with common difference $6.$ Therefore, $2016$ is the $335$ th term of this, but including $1,$ it is $336\cdot6+1=\boxed{2017}.$
|
B
|
2017
|
e3f3c2d6b4b0f37c242b48599e9c980e
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21
|
Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
|
Let $d$ be the next divisor written to the right of $323.$
If $\gcd(323,d)=1,$ then \[n\geq323d>323^2>100^2=10000,\] which contradicts the precondition that $n$ is a $4$ -digit number.
It follows that $\gcd(323,d)>1.$ Since $323=17\cdot19,$ the smallest possible value of $d$ is $17\cdot20=\boxed{340}(323,d)=17\cdot19\cdot20=6460.\] ~MRENTHUSIASM ~tdeng
|
C
|
340
|
e3f3c2d6b4b0f37c242b48599e9c980e
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21
|
Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
|
Let $d$ be the next divisor written to the right of $323.$
Since $n$ is even and $323=17\cdot19,$ we have $n=2\cdot17\cdot19\cdot k=646k$ for some positive integer $k.$ Moreover, since $1000\leq n\leq9998,$ we get $2\leq k\leq15.$ As $d>323,$ it is clear that $d$ must be divisible by $17$ or $19$ or both.
Therefore, the smallest possible value of $d$ is $17\cdot20=\boxed{340}(323,d)=17\cdot19\cdot20=6460.\] ~MRENTHUSIASM ~bjhhar
|
C
|
340
|
e3f3c2d6b4b0f37c242b48599e9c980e
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21
|
Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
|
The prime factorization of $323$ is $17 \cdot 19$ . Our answer must be a multiple of either $17$ or $19$ or both. Since $17 < 19$ , the next smallest divisor that is divisble by $17$ would be $323 + 17 = \boxed{340}$
|
C
|
340
|
e3f3c2d6b4b0f37c242b48599e9c980e
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21
|
Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
|
Since prime factorizing $323$ gives you $17 \cdot 19$ , the desired answer needs to be a multiple of $17$ or $19$ , this is because if it is not a multiple of $17$ or $19$ $n$ will be more than a $4$ digit number. For example, if the answer were to instead be $324$ $n$ would have to be a multiple of $2^2\cdot3^4\cdot17\cdot19$ for both $323$ and $324$ to be a valid factor, meaning $n$ would have to be at least $104652$ , which is too big. Looking at the answer choices, $\textbf{(A)}$ and $\textbf{(B)}$ are both not a multiple of neither $17$ nor $19$ $\textbf{(C)}$ is divisible by $17$ $\textbf{(D)}$ is divisible by $19$ , and $\textbf{(E)}$ is divisible by both $17$ and $19$ . Since $\boxed{340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$ , a $4$ -digit number. Note that $n$ is also divisible by $2$ , one of the listed divisors of $n$ . (If $n$ was not divisible by $2$ , we would need to look for a different divisor.)
|
C
|
340
|
e3f3c2d6b4b0f37c242b48599e9c980e
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_21
|
Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
|
Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$ -digit $n$ . So, we need a choice that shares a factor(s) with $323$ , such that the factors we'll need to add to the prime factorization of $n$ (in result to adding the chosen divisor) won't cause our number to multiply to more than $4$ digits.
The prime factorization of $323$ is $17\cdot19$ , and since we know $n$ is even, our answer needs to be
We see $340$ achieves this and is the smallest to do so ( $646$ being the other). So, we get $\boxed{340}$
|
C
|
340
|
5f1aca318d71d7f810bc0a6fc080121c
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_23
|
How many ordered pairs $(a, b)$ of positive integers satisfy the equation \[a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),\] where $\text{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$ , and $\text{lcm}(a,b)$ denotes their least common multiple?
$\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}$
|
Let $x = \text{lcm}(a, b)$ , and $y = \text{gcd}(a, b)$ . Therefore, $a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y$ . Thus, the equation becomes
\[x\cdot y + 63 = 20x + 12y\] \[x\cdot y - 20x - 12y + 63 = 0\]
Using Simon's Favorite Factoring Trick , we rewrite this equation as
\[(x - 12)(y - 20) - 240 + 63 = 0\] \[(x - 12)(y - 20) = 177\]
Since $177 = 3\cdot 59$ and $x > y$ , we have $x - 12 = 59$ and $y - 20 = 3$ , or $x - 12 = 177$ and $y - 20 = 1$ . This gives us the solutions $(71, 23)$ and $(189, 21)$ . Since the $\text{GCD}$ must be a divisor of the $\text{LCM}$ , the first pair does not work. Assume $a>b$ . We must have $a = 21 \cdot 9$ and $b = 21$ , and we could then have $a<b$ , so there are $\boxed{2}$ solutions.
(awesomeag)
|
B
|
2
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.
[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]
Now notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$ . We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{199}$
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Same as the first solution, $x^2=10,000\{x\}$
We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in $\{x\}$ in terms of $\lfloor x \rfloor$ \[\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0\]
We use the quadratic formula to solve for $\{x\}$ \[\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2}\]
Since $0 \leq \{x\} < 1$ , we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$
Solving over the integers, $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$ .
We can then rewrite the problem below:
$(a+k)^2 + 10000a = 10000(a+k)$
From here, we get
$(a+k)^2 + 10000a = 10000a + 10000k$
Solving for $a+k = x$
$(a+k)^2 = 10000k$
$x = a+k = \pm100\sqrt{k}$
Because $0 \leq k < 1$ , we know that $a+k$ cannot be less than or equal to $-100$ nor greater than or equal to $100$ . Therefore:
$-99 \leq x \leq 99$
There are $199$ elements in this range, so the answer is $\boxed{199}$
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Notice the given equation is equivalent to $(\lfloor x \rfloor+\{x\})^2=10,000\{x\}$
Now we know that $\{x\} < 1$ so plugging in $1$ for $\{x\}$ we can find the upper and lower bounds for the values.
$(\lfloor x \rfloor +1)^2 = 10,000(1)$
$(\lfloor x \rfloor +1) = \pm 100$
$\lfloor x \rfloor = 99, -101$
And just like $\textbf{Solution 2}$ , we see that $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Firstly, if $x$ is an integer, then $10,000\lfloor x \rfloor=10,000x$ , so $x$ must be $0$
If $0<x<1$ , then we know the following:
$0<x^2<1$
$10,000\lfloor x \rfloor =0$
$0<10,000x<10,000$
Therefore, $0<x^2+10,000\lfloor x \rfloor <1$ , which overlaps with $0<10,000x<10,000$ . This means that there is at least one real solution between $0$ and $1$ . Since $x^2+10,000\lfloor x \rfloor$ increases quadratically and $10,000x$ increases linearly, there is only one solution for this case.
Similarly, if $1<x<2$ , then we know the following:
$1<x^2<4$
$10,000\lfloor x \rfloor =10,000$
$<10,000<10,000x<20,000$
By following similar logic, we can find that there is one solution between $1$ ad $2$
We can also follow the same process to find that there are negative solutions for $x$ as well.
There are not an infinite amount of solutions, so at one point there will be no solutions when $n<x<n+1$ for some integer $n$ . For there to be no solutions in a given range means that the range of $10,000\lfloor x \rfloor + x^2$ does not intersect the range of $10,000x$ $x^2$ will always be positive, and $10,000\lfloor x \rfloor$ is less than $10,000$ less than $10,000x$ , so when $x^2 >= 10,000$ , the equation will have no solutions. This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{199}$ solutions.
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
General solution to this type of equation $f(x, \lfloor x \rfloor) = 0$
$x^2 - 10000x + 10000 \lfloor x \rfloor =0$
$x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}$ $\lfloor x \rfloor \le 2500$
$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$
If $x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}$ $x \ge 5000$ , it contradicts $x < \lfloor x \rfloor + 1 \le 2501$
So $x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}$
Let $k = \lfloor x \rfloor$ $x= 5000 - 100 \sqrt{2500 - k}$
$k \le 5000 - 100 \sqrt{2500 - k} < k + 1$
$0 \le 5000 - k - 100 \sqrt{2500 - k} < 1$
$0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1$
$0 \le (\sqrt{2500 - k} - 50)^2 < 1$
$-1 < \sqrt{2500 - k} - 50 < 1$
$49 < \sqrt{2500 - k} < 51$
$-101 < k < 99$
So the number of $k$ 's values is $99-(-101)-1=199$ . Because $x=5000-100\sqrt{2500-k}$ , for each value of $k$ , there is a value for $x$ . The answer is $\boxed{199}$
|
C
|
199
|
d63c5dd99ac37b9eba3e8c4bbabbce86
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25
|
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Subtracting $10000\lfloor x\rfloor$ from both sides gives $x^2=10000(x-\lfloor x\rfloor)=10000\{x\}$ . Dividing both sides by $10000$ gives $\left(\frac{x}{100}\right)^2=\{x\}<1$ $\left(\frac{x}{100}\right)^2<1$ when $-100<x<100$ so the answer is $\boxed{199}$
| null |
199
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: \[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{127}\]
|
C
|
127
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$ . This is always $1$ less than a power of $2$ . The only admissible answer choice by this rule is thus $\boxed{127}$
|
C
|
127
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{127}$
|
C
|
127
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
If you distribute this you get a sum of the powers of $2$ . The largest power of $2$ in the series is $64$ , so the sum is $\boxed{127}$
|
C
|
127
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
$(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ $=(2(2(2(2(2(3)+1)+1)+1)+1)+1)$ $=(2(2(2(2(6+1)+1)+1)+1)+1)$ $=(2(2(2(2(7)+1)+1)+1)+1)$ $=(2(2(2(14+1)+1)+1)+1)$ $=(2(2(2(15)+1)+1)+1)$ $=(2(2(30+1)+1)+1)$ $=(2(2(31)+1)+1)$ $=(2(62+1)+1)$ $=(2(63)+1)$ $=(126+1)$ $=127 \Longrightarrow \boxed{127}$
|
C
|
127
|
61fef4a698aff81fa7aec143f49a756d
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1
|
What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$
|
Notice that $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1$ . Substituting $2$ for $x$ , we get \[2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \Longrightarrow \boxed{127}\]
|
C
|
127
|
58304d1662f2cefaa5c879f993887488
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_2
|
Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$ -popsicle boxes for $$2$ each, and $5$ -popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$
|
$$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$ , we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ popsicles, so the answer is $\boxed{13}$
|
D
|
13
|
625849de90f117324bb7ec2dc39087ed
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_3
|
Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
[asy] draw((0,0)--(0,10)--(15,10)--(15,0)--cycle); fill((0,0)--(0,10)--(15,10)--(15,0)--cycle, lightgray); draw((1,1)--(1,3)--(7,3)--(7,1)--cycle); fill((1,1)--(1,3)--(7,3)--(7,1)--cycle, white); draw((1,4)--(1,6)--(7,6)--(7,4)--cycle); fill((1,4)--(1,6)--(7,6)--(7,4)--cycle, white); draw((1,7)--(1,9)--(7,9)--(7,7)--cycle); fill((1,7)--(1,9)--(7,9)--(7,7)--cycle, white); draw((8,1)--(8,3)--(14,3)--(14,1)--cycle); fill((8,1)--(8,3)--(14,3)--(14,1)--cycle, white); draw((8,4)--(8,6)--(14,6)--(14,4)--cycle); fill((8,4)--(8,6)--(14,6)--(14,4)--cycle, white); draw((8,7)--(8,9)--(14,9)--(14,7)--cycle); fill((8,7)--(8,9)--(14,9)--(14,7)--cycle, white); defaultpen(fontsize(8, lineskip=1)); label("2", (1.2, 2)); label("6", (4, 1.2)); defaultpen(linewidth(.2)); draw((0,8)--(1,8), arrow=Arrows); draw((7,8)--(8,8), arrow=Arrows); draw((14,8)--(15,8), arrow=Arrows); draw((11,0)--(11,1), arrow=Arrows); draw((11,3)--(11,4), arrow=Arrows); draw((11,6)--(11,7), arrow=Arrows); label("1", (.5,7.8)); label("1", (7.5,7.8)); label("1", (14.5,7.8)); label("1", (10.8,.5)); label("1", (10.8,3.5)); label("1", (10.8,6.5)); [/asy]
$\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$
|
Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.
Since the width of Tamara's garden contains three margins, the total width is $2\cdot 6+3\cdot 1 = 15$ feet.
Similarly, the height of Tamara's garden is $3\cdot 2+4\cdot 1 = 10$ feet.
Therefore, the total area of the garden is $15\cdot 10 =150$ square feet.
Finally, since the six flower beds each have an area of $2\cdot 6 = 12$ square feet, the area we seek is $150 - 6\cdot 12$ , and our answer is $\boxed{78}$
|
B
|
78
|
625849de90f117324bb7ec2dc39087ed
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_3
|
Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
[asy] draw((0,0)--(0,10)--(15,10)--(15,0)--cycle); fill((0,0)--(0,10)--(15,10)--(15,0)--cycle, lightgray); draw((1,1)--(1,3)--(7,3)--(7,1)--cycle); fill((1,1)--(1,3)--(7,3)--(7,1)--cycle, white); draw((1,4)--(1,6)--(7,6)--(7,4)--cycle); fill((1,4)--(1,6)--(7,6)--(7,4)--cycle, white); draw((1,7)--(1,9)--(7,9)--(7,7)--cycle); fill((1,7)--(1,9)--(7,9)--(7,7)--cycle, white); draw((8,1)--(8,3)--(14,3)--(14,1)--cycle); fill((8,1)--(8,3)--(14,3)--(14,1)--cycle, white); draw((8,4)--(8,6)--(14,6)--(14,4)--cycle); fill((8,4)--(8,6)--(14,6)--(14,4)--cycle, white); draw((8,7)--(8,9)--(14,9)--(14,7)--cycle); fill((8,7)--(8,9)--(14,9)--(14,7)--cycle, white); defaultpen(fontsize(8, lineskip=1)); label("2", (1.2, 2)); label("6", (4, 1.2)); defaultpen(linewidth(.2)); draw((0,8)--(1,8), arrow=Arrows); draw((7,8)--(8,8), arrow=Arrows); draw((14,8)--(15,8), arrow=Arrows); draw((11,0)--(11,1), arrow=Arrows); draw((11,3)--(11,4), arrow=Arrows); draw((11,6)--(11,7), arrow=Arrows); label("1", (.5,7.8)); label("1", (7.5,7.8)); label("1", (14.5,7.8)); label("1", (10.8,.5)); label("1", (10.8,3.5)); label("1", (10.8,6.5)); [/asy]
$\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$
|
The long horizontal walkways have an area of $15 ft^{2}$ each, and there are $4$ of them $\longrightarrow 60 ft^{2}$
The short vertical walkways have an area of $2 ft^{2}$ each, and there are $9$ of them $\longrightarrow 18 ft^{2}$
Adding the areas together, we have $60 ft^{2} + 18 ft^{2} = 78 ft^{2} \Longrightarrow \boxed{78}$ .
~JH. L
|
B
|
78
|
998fa7866f2b20723d2306139fae8eb9
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_4
|
Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$
|
Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $\boxed{14}$ minutes.
|
B
|
14
|
0afb084c04bec3904c940ffec987abce
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5
|
The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
|
Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\frac{x}{xy}+\frac{y}{xy}=4.$
\[\frac{1}{y}+\frac{1}{x}=\boxed{4}.\]
|
C
|
4
|
0afb084c04bec3904c940ffec987abce
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5
|
The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
|
Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.
See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{4}.$
|
C
|
4
|
0afb084c04bec3904c940ffec987abce
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5
|
The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
|
Notice that from the information given above, $x+y=4xy$
Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$
We can solve this by substituting $x+y\implies 4xy$
Our answer is simply $\frac{4xy}{xy}\implies4$
Therefore, the answer is $\boxed{4}$
|
C
|
4
|
5d8ab408e0eb81734129533999de138a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_7
|
Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$
|
Let $j$ represent how far Jerry walked, and $s$ represent how far Silvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2}=0.3 \implies \boxed{30}$
|
A
|
30
|
d8b29c4e1f8f80b12b8f16e8032a0a74
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\cdot20 = 200$ . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$ , or $\binom{10}{2} = 45$ . Thus the answer is $200 + 45 = \boxed{245}$
|
B
|
245
|
d8b29c4e1f8f80b12b8f16e8032a0a74
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
We can also use complementary counting. First of all, $\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\dbinom{20}{2}=190$ hugs. $435-190= \boxed{245}$
|
B
|
245
|
d8b29c4e1f8f80b12b8f16e8032a0a74
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
We can focus on how many handshakes the $10$ people who don't know anybody get.
The first person gets $29$ handshakes with other people not him/herself, the second person gets $28$ handshakes with other people not him/herself and not the first person, ..., and the tenth receives $20$ handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:
$\frac{10(20+29)}{2}\implies 5(49)\implies 245.$ Therefore, the answer is $\boxed{245}$
|
B
|
245
|
d8b29c4e1f8f80b12b8f16e8032a0a74
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
First, we can find out the number of handshakes that the $10$ people who don't know anybody share with the $20$ other people. This is simply $10 \cdot 20 = 200$ . Next, we need to find out the number of handshakes that are shared within the $10$ people who don't know anybody. Here, we can use the formula $\frac{n(n-1)}{2}$ , where $n$ is the number of people being counted. The reason we divide by $2$ is because $n(n-1)$ counts the case where the $1^{st}$ person shakes hands with the $2^{nd}$ person $and$ the case where the $2^{nd}$ shakes hands with the $1^{st}$ (and these 2 cases are the same). Thus, plugging $n=10$ gives us $\frac{10 \cdot 9}{2} \implies 45$ . Adding up the 2 cases gives us $200+45=\boxed{245}$
|
B
|
245
|
007d2970a5423a9fbbb05dee9c37fb91
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_9
|
Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$ , a distance of $10$ km all uphill, then from town $B$ to town $C$ , a distance of $15$ km all downhill, and then back to town $A$ , a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$ -km ride than it takes Penny?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95$
|
The distance from town $A$ to town $B$ is $10$ km uphill, and since Minnie rides uphill at a speed of $5$ kph, it will take her $2$ hours. Next, she will ride from town $B$ to town $C$ , a distance of $15$ km all downhill. Since Minnie rides downhill at a speed of $30$ kph, it will take her half an hour. Finally, she rides from town $C$ back to town $A$ , a flat distance of $20$ km. Minnie rides on a flat road at $20$ kph, so this will take her $1$ hour. Her entire trip takes her $3.5$ hours.
Secondly, Penny will go from town $A$ to town $C$ , a flat distance of $20$ km. Since Penny rides on a flat road at $30$ kph, it will take her $\frac{2}{3}$ of an hour. Next Penny will go from town $C$ to town $B$ , which is uphill for Penny. Since Penny rides at a speed of $10$ kph uphill, and town $C$ and $B$ are $15$ km apart, it will take her $1.5$ hours. Finally, Penny goes from Town $B$ back to town $A$ , a distance of $10$ km downhill. Since Penny rides downhill at $40$ kph, it will only take her $\frac{1}{4}$ of an hour. In total, it takes her $29/12$ hours, which simplifies to $2$ hours and $25$ minutes.
Finally, Penny's $2$ hr $25$ min trip was $\boxed{65}$ minutes less than Minnie's $3$ hr $30$ min trip.
|
C
|
65
|
a8cbdd23667266dbd0217a818f8b3cdd
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_10
|
Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$
|
The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{17}$
|
B
|
17
|
f98a768aeea7f6f5ff34a9d001e0b7b5
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_11
|
The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
|
In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$ ):
$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$ , where $x$ is equal to the length of our line segment.
Solving, we find that $x = \boxed{20}$
|
D
|
20
|
f98a768aeea7f6f5ff34a9d001e0b7b5
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_11
|
The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
|
Because this is just a cylinder and $2$ hemispheres ("half spheres"), and the radius is $3$ , the volume of the $2$ hemispheres is $\frac{4(3^3)\pi}{3} = 36 \pi$ . Since we also know that the volume of this whole thing is $216 \pi$ , we do $216-36$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$ , so our answer is $\boxed{20}.$
|
D
|
20
|
3b0dce925c663ecf8ab86a7fc41df98b
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_13
|
Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$
|
A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\boxed{9}$
|
D
|
9
|
2492b7149b558e19021bfef9582f86dc
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14
|
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) } 23\%\qquad \textbf{(E) } 25\%$
|
Let $m$ = cost of movie ticket Let $s$ = cost of soda
We can create two equations:
\[m = \frac{1}{5}(A - s)\] \[s = \frac{1}{20}(A - m)\]
Substituting we get:
\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\] which yields: \[m = \frac{19}{99}A\]
Now we can find s and we get:
\[s = \frac{4}{99}A\]
Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{23}\]
|
D
|
23
|
2492b7149b558e19021bfef9582f86dc
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14
|
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) } 23\%\qquad \textbf{(E) } 25\%$
|
We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$ .
Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$ .
Adding, we get $\frac{2300}{99}$ , which is closest to $23$ which is $\boxed{23}$
|
D
|
23
|
2492b7149b558e19021bfef9582f86dc
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14
|
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) } 23\%\qquad \textbf{(E) } 25\%$
|
Let $m$ be the price of a movie ticket and $s$ be the price of a soda.
Then,
\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]
Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$ . Assume WLOG that $s=4$ $m=19$ , thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$ . Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{23}$
|
D
|
23
|
bdbd65285ddefe04bdcd50450bddc592
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16
|
There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$
|
If we have horses, $a_1, a_2, \ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12$ . Finally, $1+2 = \boxed{3}$
|
B
|
3
|
bdbd65285ddefe04bdcd50450bddc592
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16
|
There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$
|
We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$ $2$ $3$ , or $4$ . We quickly consider $12$ since it is the smallest number that is the LCM of $1$ $2$ $3$ and $4$ . Since $12$ has $5$ single-digit divisors, namely $1$ $2$ $3$ $4$ , and $6$ , our answer is $1+2 = \boxed{3}$
|
B
|
3
|
bdbd65285ddefe04bdcd50450bddc592
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16
|
There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$
|
First, for 5 horses to simultaneously pass the starting line after $T$ seconds, $T$ must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of $T$ , and therefore meaning $T$ must have at least $5$ $1$ -digit divisors. Since we want to minimize $T$ , we will start by guessing the lowest natural number, $1$ $1$ has only $1$ factor, so it does not work, we now repeat the process for the numbers between $2$ and $12$ (This should not take more than a minute) to get that $12$ is the first number to have $5$ or more single-digit divisors ( $1, 2, 3, 4, 6$ ). The sum of the digits of $12$ is $1+2 = \boxed{3}$ , which is our answer.
|
B
|
3
|
bdbd65285ddefe04bdcd50450bddc592
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16
|
There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$
|
By inspection, $(1, 2, 3, 4, 6)$ yields the lowest answer of $12$ and the sum of the digits is $1+2 \Longrightarrow \boxed{3}$
|
B
|
3
|
77b5c97a3323e1f938f3ec5baee7f3c4
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_17
|
Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$
|
Because $P$ $Q$ $R$ , and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational. The greatest value of $PQ$ happens when $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. Another possible pair could be $(-4,3)$ and $(5,0)$ . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from $(-4,3)$ to $(4,-3)$ . The distance between $(3,-4)$ and $(-4,3)$ is greater than the distance between $(5,0)$ and $(4,-3)$ . Therefore, the segment from $(3,-4)$ to $(-4,3)$ is the longest attainable (the other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$ $(3, 4)$ and $(-4, -3)$ $(-3, 4)$ and $(4, -3)$ . The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3)$ and vice versa. Using the distance formula, we get that $PQ$ is $\sqrt{98}$ and that $RS$ is $\sqrt{2}.$ $\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{7}$
|
D
|
7
|
77b5c97a3323e1f938f3ec5baee7f3c4
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_17
|
Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$
|
We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using $7$ (though $5 \sqrt{2}$ actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let $RS$ be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of $RS$ is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$ . Assuming that $\frac{PQ}{RS} = 7$ , we plug in $RS = \sqrt{2}$ and solve for $PQ$ $PQ=7\sqrt{2}$ . Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of $x$ and $y$ does $\sqrt{x^{2}+y^{2}}=7\sqrt2$ ? We see that this can easily be made into a $45-45-90$ triangle. But, instead of substituting $y=x$ into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a $45-45-90$ triangle, the two $45$ degree sides have side length $s$ , then the hypotenuse is $s\sqrt2$ . Using this, we can see that $s=7$ , and since our equation does in fact yield a sensible solution, we can be assured that our answer is $\boxed{7}$
|
D
|
7
|
77b5c97a3323e1f938f3ec5baee7f3c4
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_17
|
Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$
|
By inspection, when $R$ is at $(3, 4)$ and $S$ is at $(4, 3),$ it makes $RS$ as small as possible with a distance of $\sqrt{2}$ . The greatest possible length of $PQ$ arises when $P$ is at $(-3, 4)$ and $Q$ is at $(4, -3).$ Using the distance formula, we find that $PQ$ has a length of $7\sqrt{2}.$ The requested fraction is then $\dfrac{PQ}{RS} = \dfrac{7\sqrt{2}}{\sqrt{2}} = \boxed{7}$
|
D
|
7
|
ec2dbb5106882670665f78b963a6040c
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_18
|
Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$ , since if she gets to her turn again, she is back where she started with probability of winning $P$ . The chance she wins on her first turn is $\frac{1}{3}$ . The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$ . Multiplying gives us $\frac{2}{5}$ . Thus, \[P = \frac{1}{3} + \frac{2}{5}P\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5=\boxed{4}$
|
D
|
4
|
ec2dbb5106882670665f78b963a6040c
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_18
|
Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} \cdots$ This can be represented as an infinite geometric series: \[P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5 = \boxed{4}.$
|
D
|
4
|
ec2dbb5106882670665f78b963a6040c
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_18
|
Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
We can solve this by using 'casework,' the cases being:
Case 1: Amelia wins on her first turn.
Case 2 Amelia wins on her second turn.
and so on.
The probability of her winning on her first turn is $\dfrac13$ . The probability of all the other cases is determined by the probability that Amelia and Blaine all lose until Amelia's turn on which she is supposed to win. So, the total probability of Amelia winning is: \[\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)\cdot\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)^2\cdot\dfrac{1}{3}+\cdots.\] Factoring out $\dfrac13$ we get a geometric series: \[\dfrac{1}{3}\left(1+\dfrac{2}{5}+\left(\dfrac{2}{5}\right)^2+\cdots\right) = \dfrac{1}{3}\cdot\dfrac{1}{3/5} = \boxed{59}.\]
| null |
59
|
05c9f66d5524610e20b9110501cf3f02
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19
|
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$
|
Let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.
We can split this problem up into two cases:
$\textbf{Case 1: }$ A sits on an edge seat.
Since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of $2 \cdot 2 \cdot 2 \cdot 2 = 16$
$\textbf{Case 2: }$ A does not sit in an edge seat.
Still, the only two people that can sit next to A are either D or E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are $3 \cdot 2 \cdot 2 = 12$ seatings in this case.
Adding up all the seatings, we have $16+12 = \boxed{28}$
|
C
|
28
|
05c9f66d5524610e20b9110501cf3f02
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19
|
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$
|
Label the seats (from left to right) $1$ through $5$ . The number of ways to seat Derek and Eric in the five seats with no restrictions is $5 \cdot 4=20$ . The number of ways to seat Derek and Eric such that they sit next to each other is $8$ (we can treat Derek and Eric as a "block". There are four ways to seat this "block", and two ways to permute Derek and Eric, for a total of $4\cdot 2=8$ ), so the number of ways such that Derek and Eric don't sit next to each other is $20-8=12$ . Note that once Derek and Eric are seated, we can divide into three cases.
The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us $0$ ways.
Another possible case is if Derek and Eric sit in seats $2$ and $4$ in some order. There are $2$ possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are $3!=6$ ways to do this. So the second case gives us $2 \cdot 6=12$ total ways.
The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are $12-2-2=8$ ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us $8 \cdot 2=16$ ways.
So in total there are $12+16=28$ ways. Our answer is $\boxed{28}$ Minor $\LaTeX$ edits by fasterthanlight
|
C
|
28
|
05c9f66d5524610e20b9110501cf3f02
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19
|
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$
|
We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, than when they aren't.
PIE:
Let's count the total number of cases where one of these is true:
When Alice is with Bob: $2\cdot4!=48$
When Alice is with Carla: $2\cdot4!=48$
When Derek is with Eric: $2\cdot4!=48$
Then, we count the cases where two of these are true.
Alice is next to Carla, and Alice is also next to Bob.
There are two ways to rearrange Alice, Bob, and Carla so that this is true:
BAC and CAB. $2\cdot3!=12$
Alice is next to Carla, and Derek is also next to Eric. $2\cdot2\cdot3!=24$
Alice is next to Bob, and Derek is also next to Eric. $2\cdot2\cdot3!=24$
Finally, we count the cases where all three of these are true: $2\cdot2\cdot2=8$
We add up the cases where one of these are true: $48\cdot3=144$
Subtract the cases where two of these are true: $144-60=84$
And finally add back the cases where three of these are true: $84+8=92$
Thus, our answer is $5!-92=28$ , or $\boxed{28}$
|
C
|
28
|
05c9f66d5524610e20b9110501cf3f02
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19
|
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$
|
To find the number of ways, we do casework.
Case 1: Alice sits in the first seat (leftmost)
Since Alice refuses to sit with Bob and Carla, then the seat on her immediate right must be Derek or Eric. The middle seat must be Bob or Carla (because Derek and Eric refuse to sit together). The seat to the right of the middle seat could be whoever is left over from Derek and Eric, or whoever is left together from Bob and Carla. The last seat only has one person left. There are $4$ ways to permute Bob, Carla, Derek, and Eric, and $2$ ways to pick who goes in the seat to the right of the middle seat, for $4\cdot2=8$ seating arrangements here.
Case 2: Alice sits in the second seat
Derek and Eric must be on both sides of Alice because otherwise, we would have to put Bob or Carla next to Alice which is forbidden. Then Bob and Carla take the remaining two seats. There are $4$ ways to permute and $4$ arrangements here.
Case 3: Alice sits in the middle seat
Once again, Derek and Eric must be on both sides of Alice, and Bob and Carla need to take the two remaining seats. There are also $4$ ways to permute and $4$ arrangements here.
Case 4: Alice sits in the fourth seat
By symmetry, this is the same as case 2. There are $4$ arrangements.
Case 5: Alice sits in the last seat (rightmost)
By symmetry, this is the same as case 1. There are $8$ arrangements.
Adding up the cases, there are $8+4+4+4+8=28$ total seating arrangements $\Longrightarrow \boxed{28}$
|
C
|
28
|
6447eec31927b86db43e3c0707b7ea23
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_20
|
Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$
|
Note that $n \equiv S(n) \pmod{9}$ . This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$ . Thus, if $S(n) = 1274$ , then $n \equiv 5 \pmod{9}$ , and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that satisfies $n+1 \equiv 6 \pmod{9}$ is $\boxed{1239}$
|
D
|
1239
|
6447eec31927b86db43e3c0707b7ea23
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_20
|
Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$
|
One divisibility rule that we can use for this problem is that a multiple of $9$ will always have its digits sum to a multiple of $9$ . We can find out that the least number of digits the number $n$ has is $142$ , with $141$ $9$ 's and $1$ $5$ , assuming the rule above. No matter what arrangement or different digits we use, the divisibility rule stays the same. To make the problem simpler, we can just use the $141$ $9$ 's and $1$ $5$ . By randomly mixing the digits up, we are likely to get: $9999$ ... $9995999$ ... $9999$ . By adding $1$ to this number, we get: $9999$ ... $9996000$ ... $0000$ .
Knowing that $n+1$ is divisible by $9$ when $6$ , we can subtract $6$ from every available choice, and see if the number is divisible by $9$ afterwards.
After subtracting $6$ from every number, we can conclude that $1233$ (originally $1239$ ) is the only number divisible by $9$ .
So our answer is $\boxed{1239}$
|
D
|
1239
|
6447eec31927b86db43e3c0707b7ea23
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_20
|
Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$
|
The number $n$ can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.
If $A$ is correct, then $n$ must be some number $99999999...9$ , because when we add one to $99999999...9$ we get $10000000...00$ . Thus, if $1$ is the correct answer, then the equation $9x=1274$ must have an integer solution (i.e. $1274$ must be divisible by $9$ ). But since it does not, $1$ is not the correct answer.
If $B$ is correct, then $n$ must be some number $29999999...9$ , because when we add one to $29999999...9$ , we get $30000000...00$ . Thus, if $3$ is the correct answer, then the equation $2+9x=1274$ must have an integer solution. But since it does not, $3$ is not the correct answer.
Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if $S(n+1)=N$ , then $n$ must be a number whose initial digits sum to $N-1$ , and whose other, terminating digits, are all $9$ . Thus, we can evaluate the three final possibilities by seeing if the equation $(N-1)+9x=1274$ has an integer solution.
The equation does not have an integer solution for $N=12$ , so $C$ is not correct. However, the equation does have an integer solution for $N=1239$ $x=4$ ), so $\boxed{1239}$ is the answer.
|
D
|
1239
|
6447eec31927b86db43e3c0707b7ea23
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_20
|
Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$
|
If adding $1$ to $n$ does not carry any of its digits, then $S(n+1)=S(n)+1$ (ex: $25+1=26$ . Sum of digits $7 \rightarrow 8$ ). But since no answer choice is $1275$ , that means $n$ has some amount of $9$ 's from right to left.
When $n+1$ , some $9$ 's will bump to 0, not affected its $\pmod 9$ . But the first non-9 digit (from right to left) will be bumped up by 1. So $S(n) + 1 \pmod {9} \equiv S(n+1) \pmod{9}$ . For example, $34999+1=35000$ , and the sum of digits $7+27 \rightarrow 8+0$
Since $S(n) \equiv 5 \pmod{9}$ , that means $S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that meets this requirement is $\boxed{1239}.$
|
D
|
1239
|
36aae137946320e95f59d29cddbac1c0
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_23
|
How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$
|
We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$ , so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$ , which simplifies to $2300$ .
Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\dbinom53$ is $10$ , and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$ .
We can also count the ones with $4$ on a diagonal. That is $\dbinom43$ , which is 4, and there are $4$ of those diagonals, so that results in $16$ .
We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$ .
We can also count the ones with a slope of $\frac12$ $2$ $-\frac12$ , or $-2$ , with $3$ points in each. Note that there are $3$ such lines, for each slope, present in the grid. In total, this results in $12$ .
Finally, we subtract all the ones in a line from $2300$ , so we have $2300-120-16-4-12=\boxed{2148}$
|
B
|
2148
|
36aae137946320e95f59d29cddbac1c0
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_23
|
How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$
|
There are $5 \times 5 = 25$ total points in all. So, there are $\dbinom{25}3 = 2300$ ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line.
There are $10 \times 10 = 100$ cases where the 3 points chosen make up a vertical or horizontal line.
There are $2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40$ cases where the 3 points all land on the diagonals of the square.
There are $3 \times 4=12$ ways where the 3 points make the a slope of $\frac{1}{2}$ $-\frac{1}{2}$ $2$ , and $-2$
Hence, there are $100+40+12=152$ cases where the chosen 3 points make a line. The answer would be $2300-152=\boxed{2148}$
|
B
|
2148
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
$f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x+p)\]
where $-p\in\mathbb{R}$ is the fourth root of $f(x)$ .
(Using $(x+p) = (x-r))$ instead of $(x-r)$ makes the following computations less messy.)
Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x+p)\\ &=x^4+(a+p)x^3+(1+ap)x^2+(10+p)x+10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a+p&=1\\ 1+ap=b\\ 10+p&=100\\ 10p&=c.\\ \end{align*}
Let's solve for $a,b,c,$ and $p$ . Since $10+p=100$ $p=90$
Since $a+p=1$ $a=-89$
(Solution 1.1 branches from here and takes a shortcut.)
$c=(10)(90)=900$
Then, since $b=1+ap$ $b=-8009$ . Thus,
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
(Solution 1.2 branches from here and takes another shortcut)
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007}
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
A faster ending to Solution 1 is as follows.
\begin{align*} f(1)&=(1+p)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &= (7)(13)(11)(-7) = (1001)(-7) \\ &=\boxed{7007}
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
Also a faster ending to Solution 1 is as follows.
To find $f(1)$ we just need to find the sum of the coefficients which is $1 + 1 - 8009 + 100 + 900= \boxed{7007}.$
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored form:
\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})\]
We can expand the expression on the right-hand side to get:
\[f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c\]
Now we have $f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c$
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]
and finally,
\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\]
We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$ . We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{7007}$
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3 & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{7007}$
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
Let the roots of $g(x)$ be $r_1$ $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ $r_2$ $r_3$ , and $r_4$ . From Vieta's, we have: \begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*} The fourth root is $a-1$ . Since $r_1$ $r_2$ , and $r_3$ are common roots, we have: \begin{align*} f(x)=g(x)(x-(a-1)) \\ f(1)=g(1)(1-(a-1)) \\ f(1)=(a+12)(2-a) \\ f(1)=-(a+12)(a-2) \\ \end{align*} Let $a-2=k$ \begin{align*} f(1)=-k(k+14) \end{align*} Note that $-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)$ This gives us a pretty good guess of $\boxed{7007}$
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ \[\therefore 90=1-a.\] The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ $\dfrac{b-1}{1-a} = a$ \[\therefore b-1=a(1-a)=-89*90=-8010\] and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ \[\therefore c=10(1-a)=10*90=900\] Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{7007}$
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after we multiply $f(x)$ by $(1-a)$ . Now we equate coefficients of same-degree $x$ terms. This gives us $10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009$ . We are interested in finding $f(1)$ , which equals $1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{7007}$ . ~skyscraper
|
C
|
7007
|
12da7c46dcc398f0d21c4c03ab49a91e
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24
|
For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
|
We first note that $f(x) = g(x) \cdot q(x) + r(x)$ where $q$ is the quotient function and $r$ is the remainder function.
Clearly, $r(x) = 0$ because every single root in $g$ is also in $f$ , thus implying $g$ divides $f$ .
So, we wish to find $f(1) = g(1) \cdot q(1)$
Such an expression for $g(1)$ is pretty clean here as we can obtain $g(1) = a + 12$ , so we rewrite $f(1) = (a + 12) \cdot q(1)$ .
Well, now we need to know how $q$ is expressed in order to obtain $q(1)$ . This motivates us to long divide to obtain the quotient function. After simple long division $q(x) = x + (1 - a)$ . In addition, what is left over, namely $r(x)$ , has a constant piece of $a + 89$ (you'll see in a few sentences why we only care about particularly the constant piece).
Now we can write: $f(1) = (a + 12) \cdot (2 - a)$
Now, as we have already established $r(x) = 0$ for ALL $x$ that means $r(0)$ or the constant piece is $0$ , so $89 + a = 0$ , in which we obtain $a = -89$ . We now plug this back into our equation for $f(1)$ to get $(-89 + 12)(2 - (89)) = -77 \cdot 91 = \boxed{7007}$ . ~triggod
|
C
|
7007
|
936651cc1bcb35d2e24c382c6cd0cd38
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_25
|
How many integers between $100$ and $999$ , inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486$
|
There are 81 multiples of 11 between $100$ and $999$ inclusive. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Switching shows we have overcounted by a factor of 2, so assign $6 \div 2 = 3$ permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have $0$ as a digit. Since $0$ cannot be the digit of the hundreds place, we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243) to get 226. $\boxed{226}$
|
A
|
226
|
936651cc1bcb35d2e24c382c6cd0cd38
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_25
|
How many integers between $100$ and $999$ , inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486$
|
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same.
By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
$\textbf{Case 2a:}$ There are $8$ multiples of $11$ without a zero that have this property: $121$ $242$ $363$ $484$ $616$ $737$ $858$ $979$ .
Each contributes $3$ valid permutations, so there are $8 \cdot 3 = 24$ permutations in this subcase.
$\textbf{Case 2b:}$ There are $9$ multiples of $11$ with a zero that have this property: $110$ $220$ $330$ $440$ $550$ $660$ $770$ $880$ $990$ .
Each one contributes $2$ valid permutations (the first digit can't be zero), so there are $9 \cdot 2 = 18$ permutations in this subcase.
$\textbf{Case 3:}$ All the digits are different.
Since there are $\frac{990-110}{11}+1 = 81$ multiples of $11$ between $100$ and $999$ , there are $81-8-9 = 64$ multiples of $11$ remaining in this case. However, $8$ of them contain a zero, namely $209$ $308$ $407$ $506$ $605$ $704$ $803$ , and $902$ . Each of those multiples of $11$ contributes $2 \cdot 2=4$ valid permutations, but we overcounted by a factor of $2$ ; every permutation of $209$ , for example, is also a permutation of $902$ . Therefore, there are $8 \cdot 4 / 2 = 16$ . Therefore, there are $64-8=56$ remaining multiples of $11$ without a $0$ in this case. Each one contributes $3! = 6$ valid permutations, but once again, we overcounted by a factor of $2$ (note that if a number ABC is a multiple of $11$ , then so is CBA). Therefore, there are $56 \cdot 6 / 2 = 168$ valid permutations in this subcase.
Adding up all the permutations from all the cases, we have $24+18+16+168 = \boxed{226}$
|
A
|
226
|
936651cc1bcb35d2e24c382c6cd0cd38
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_25
|
How many integers between $100$ and $999$ , inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486$
|
We can first overcount and then subtract.
We know that there are $81$ multiples of $11$
We can then multiply by $6$ for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)
Now divide by $2$ , because if a number $abc$ with digits $a$ $b$ , and $c$ is a multiple of $11$ , then $cba$ is also a multiple of $11$ so we have counted the same permutations twice.
Basically, each multiple of $11$ has its own $3$ permutations (say $abc$ has $abc$ $acb$ and $bac$ whereas $cba$ has $cba$ $cab$ and $bca$ ). We know that each multiple of $11$ has at least $3$ permutations because it cannot have $3$ repeating digits.
Hence we have $243$ permutations without subtracting for overcounting.
Now note that we overcounted cases in which we have $0$ 's at the start of each number. So, in theory, we could just answer $A$ and then move on.
If we want to solve it, then we continue.
We overcounted cases where the middle digit of the number is $0$ and the last digit is $0$
Note that we assigned each multiple of $11$ three permutations.
The last digit is $0$ gives $9$ possibilities where we overcounted by $1$ permutation for each of $110, 220, ... , 990$
The middle digit is $0$ gives $8$ possibilities where we overcount by $1$ $605, 704, 803, 902$ and $506, 407, 308, 209$
Subtracting $17$ gives $\boxed{226}$
|
A
|
226
|
936651cc1bcb35d2e24c382c6cd0cd38
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_25
|
How many integers between $100$ and $999$ , inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486$
|
Taken from solution three, we notice that there are a total of $81$ multiples of $11$ between $100$ and $999$ , and each of them have at most $6$ permutation (and thus is a permutations of $6$ numbers), giving us a maximum of $486$ valid numbers.
However, if $abc$ can be divided by $11$ , so can $cba$ , which is distinct if $c \neq a$ . And if $c = a$ then $abc$ and $cba$ have the same permutations. Either way, we have doubled counted.
This reduces the number of permutations to $486/2 = 243$
Furthermore, if $a=b$ or $c=0$ (which turn out to be equivalent conditions! for example $220$ ), not all (inverse) permutations are distinct ( $\mathbf{2}20 = 2\mathbf{2}0$ ) or valid ( $022$ ). (There are 9 of these.)
Similarly, for $a0b$ , not all (inverse) permutations are valid. (There are 8 of these.)
As long as you notice at least one example of one of these 3 cases, you may infer that the answer must be smaller than $243$ . This leaves us with only one possible answer: $\boxed{226}$
|
A
|
226
|
f705fb95c47efef6ac29adb54fba744a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_1
|
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
|
Let her $2$ -digit number be $x$ . Multiplying by $3$ makes it a multiple of $3$ , meaning that the sum of its digits is divisible by $3$ . Adding on $11$ increases the sum of the digits by $1+1 = 2,$ (we can ignore numbers such as $39+11=50$ ) and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be $2$ more than a multiple of $3$ . There are two such numbers between $71$ and $75$ $71$ and $74.$ Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed: \[\] For $71,$ we reverse the digits, resulting in $17.$ Subtracting $11$ , we get $6.$ We can already see that dividing this by $3$ will not be a two-digit number, so $71$ does not meet our requirements. \[\] Therefore, the answer must be the reversed steps applied to $74.$ We have the following: \[\] $74\rightarrow47\rightarrow36\rightarrow12$ \[\] Therefore, our answer is $\boxed{12}$
|
B
|
12
|
f705fb95c47efef6ac29adb54fba744a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_1
|
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
|
Working backwards, we reverse the digits of each number from $71$ $75$ and subtract $11$ from each, so we have \[6, 16, 26, 36, 46\] The only numbers from this list that are divisible by $3$ are $6$ and $36$ . We divide both by $3$ , yielding $2$ and $12$ . Since $2$ is not a two-digit number, the answer is $\boxed{12}$
|
B
|
12
|
f705fb95c47efef6ac29adb54fba744a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_1
|
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
|
You can just plug in the numbers to see which one works. When you get to $12$ , you multiply by $3$ and add $11$ to get $47$ . When you reverse the digits of $47$ , you get $74$ , which is within the given range. Thus, the answer is $\boxed{12}$
|
B
|
12
|
f705fb95c47efef6ac29adb54fba744a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_1
|
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
|
Let x be the original number. The last digit of $3x+11$ must be $7$ so the last digit of $3x$ must be $6$ . The only answer choice that satisfies this is $\boxed{12}$
|
B
|
12
|
f705fb95c47efef6ac29adb54fba744a
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_1
|
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
|
Subtract $11$ from the numbers $71$ through $75$ . This yields $71-11 = 60$ $72-11 = 61$ $73-11 = 62$ $74-11 = 63$ , and $75-11 = 64$ . Of these, the only ones divisible by $3$ are $60$ and $63$ . Therefore, the only possible values are $71$ and $74$ . Switching the digits of each, we get $17$ and $47$ . Subtracting $11$ from each, we get the numbers $6$ and $36$ . Dividing each by $3$ , we get $2$ and $12$ . The only two-digit number is $12$ , so the answer is $\boxed{12}$
|
B
|
12
|
7c7b929da1ffdf4c179ef4c871d0f923
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_4
|
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
|
Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\implies x=y$ .
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{2}$
|
D
|
2
|
7c7b929da1ffdf4c179ef4c871d0f923
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_4
|
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
|
Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\frac{x+3y}{3x-y}=\boxed{2}$
|
D
|
2
|
7c7b929da1ffdf4c179ef4c871d0f923
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_4
|
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
|
Let $y=ax$ . The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{2}$ - mathleticguyyy
|
D
|
2
|
7c7b929da1ffdf4c179ef4c871d0f923
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_4
|
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
|
Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\boxed{2}$
|
D
|
2
|
ebcbdab40a27280404bf0c0d9b9ff113
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_5
|
Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$
|
Denote the number of blueberry and cherry jelly beans as $b$ and $c$ respectively. Then $b = 2c$ and $b-10 = 3(c-10)$ . Substituting, we have $2c-10 = 3c-30$ , so $c=20$ $b=\boxed{40}$
|
D
|
40
|
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