2018 AMC 8 Problems/Problem 7

Revision as of 06:47, 26 December 2022 by Saxstreak (talk | contribs) (Solution)

Problem

The $5$-digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution

We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for $U$ then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $\boxed{\textbf{(B) }3}$.

20187 mod 8 =_ 3

Video Solution

https://youtu.be/6xNkyDgIhEE?t=2341

https://youtu.be/doHZiAT36BY

~savannahsolver

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png