Difference between revisions of "2018 AMC 8 Problems/Problem 21"

Problem

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution 1

Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three digit≤ integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$

Solution 2

Let us create the equations: $6x+2 = 9y+5 = 11z+7$, and we know $100 \leq 11z+7 <1000$, it gives us $9 \leq z \leq 90$, which is the range of the value of z. Because of $6x+2=11z+7$, then $6x=11z+5=6z+5(z+1)$, so (z+1) must be mutiples of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of $6$ and $9$, which means multiples of $18(LCM of 6 & 9)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.). So let's say $z+1 = 18p$, then $9 \leq z = 18p-1 \leq 90, 1 \leq p \leq 91/18 or 1 \leq p \leq 5$. Thus our answer is $\boxed{\textbf{(E) }5}$ ~LarryFlora

Video Solution

https://youtu.be/CPQpkpnEuIc - Happytwin