2017 AMC 10B Problems/Problem 23

Revision as of 10:43, 16 February 2017 by Thedoge (talk | contribs) (Problem 23)

Problem 23

Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution

We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, $N \equiv 4 \text{ (mod 5)}$. The remainder when divided by 9 is $1+2+3+4 \cdot 1+0+1+1 \cdot 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 \cdot 10+11+12 \cdot 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$, which clearly has a remainder of 0 mod 9. Therefore, by CRT, the answer is $\boxed{\textbf{(C) } 9}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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 AMC 10 Problems and Solutions

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