2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 21 Hard

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?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:E

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 multiple of 6. Because of 9y+5=11z+7, then 9y=11z+2=9z+2(z+1), so (z+1) must also be a multiple 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}.