# 2017 AMC 12B Problems/Problem 19

## Problem

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 will consider this number $\bmod\ 5$ and $\bmod\ 9$. By looking at the last digit, it is obvious that the number is $\equiv 4\bmod\ 5$. To calculate the number $\bmod\ 9$, note that $$123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,$$

so it is equivalent to $$\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.$$

Let $x$ be the remainder when this number is divided by $45$. We know that $x\equiv 0 \pmod {9}$ and $x\equiv 4 \pmod {5}$, so by the Chinese remainder theorem, since $9(-1)\equiv 1 \pmod{5}$, $x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}$, or $x\equiv -36 \equiv 9 \pmod {45}$. So the answer is $\boxed {\textbf {(C)}}$.

## Solution 2

We know that this number is divisible by $9$ because the sum of the digits is $270$, which is divisible by $9$. If we subtracted $9$ from the integer we would get $1234 \cdots 4335$, which is also divisible by $5$ and by $45$. Thus the remainder is $9$, or $\boxed{\textbf{C}}$.

## Solution 3 (Beginner's Method)

To find the sum of digits of our number, we break it up into $5$ cases, starting with $0$, $1$, $2$, $3$, or $4$.

Case 1: $1+2+3+\cdots+9 = 45$,

Case 2: $1+0+1+1+1+2+\cdots+1+8+1+9 = 55$ (We add 10 to the previous cases, as we are in the next ten's place)

Case 3: $2+0+2+1+\cdots+2+9 = 65$,

Case 4: $3+0+3+1+\cdots+3+9 = 75$,

Case 5: $4+0+4+1+\cdots+4+4 = 30$,

Thus the sum of the digits is $45+55+65+75+30 = 270$, so the number is divisible by $9$. We notice that the number ends in " $4$", which is $9$ more than a multiple of $5$. Thus if we subtracted $9$ from our number it would be divisible by $5$, and $5\cdot 9 = 45$. (Multiple of n - n = Multiple of n)

So our remainder is $\boxed{\textbf{(C)}\,9}$, the value we need to add to the multiple of $45$ to get to our number.

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