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 a mutiple of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so $(z+1)$ must also be a mutiple of $9$. Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$, which means multiples of $18 (LCM \text{ of }\ 6, 9)$. So, let's say $z+1 = 18p$; then, $9 \leq z = 18p-1 \leq 90$, so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$. Thus, the answer is $\boxed{\textbf{(E) }5}$.

~LarryFlora

Solution 3

By the Chinese Remainder Theorem, we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{\textbf{(E) }5}.$

~mathboy282

Solution 4

We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent to their LCM. Thus, $x+4 \equiv 0 \mod 198,$ since 198 is the LCM of $6, 9,$ and $11.$ Since these numbers have to be three digits, they can only be $194, 392, 590, 788,$ and $986.$ This gives us the answer of $\boxed{\textbf{(E) }5}.$

~ethancui0529

Solution 5

Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$. Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$. Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$. From this, we can figure out that $N$ can be 5 different three digit numbers -- $194, 392, 590, 788,$ and $986$. $\therefore$, the answer is $\boxed{\textbf{(E) }5}.$

~DY

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 2018 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions