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

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. Therefore, by CRT, the answer is $\boxed{\textbf{(C) } 9}$ .

Note: the sum of the digits of $N$ is $270$

Solution 2

Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by $9$ . From $1$ thru $9$ , the sum is $45$ . $10$ thru $19$ , the sum is $55$ , $20$ thru $29$ is $65$ , and $30$ thru $39$ is $75$ . Thus the sum of the digits is $45+55+65+75+4+5+6+7+8 = 240+30 = 270$ , and thus $N$ is divisible by $9$ . Now, refer to the above or below solutions. $N \equiv 4 \text{ (mod 5)}$ and $N \equiv 0 \text{ (mod 9)}$ . From this information, we can conclude that $N \equiv 54 \text{ (mod 5)}$ and $N \equiv 54 \text{ (mod 9)}$ . Therefore, $N \equiv 54 \text{ (mod 45)}$ . The remainder is thus $9$ , so the answer is $\boxed{\textbf{(C) }9}$

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2017 AMC 10B Problems/Problem 23

Contents

Problem 23

Solution

Solution 2

See Also