# 2017 AMC 10B Problems/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 1

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 $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$, which has a remainder of 0 mod 9. Solving these modular congruence using CRT(Chinese Remainder Theorem) we get the remainder to be $9 \pmod{45}$. Therefore, the answer is $\boxed{\textbf{(C) } 9}$.

## Alternative Ending to Solution 1

Once we find our 2 modular congruences, we can narrow our options down to ${C}$ and ${D}$ because the remainder when $N$ is divided by $45$ should be a multiple of 9 by our modular congruence that states $N$ has a remainder of $0$ when divided by $9$. Also, our other modular congruence states that the remainder when divided by $45$ should have a remainder of $4$ when divided by $5$. Out of options $C$ and $D$, only $\boxed{\textbf{(C) } 9}$ satisfies that the remainder when $N$ is divided by 45 $\equiv 4 \text{ (mod 5)}$.

## Solution 3

Realize that $10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}$ for all positive integers $k$.

Apply this on the expanded form of $N$: $$N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv$$ $$10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv$$ $$10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}$$

## See Also

 2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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.

Invalid username
Login to AoPS