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 $\boxed {\bold {(C)}}$ .

Solution 2

We know that this number is divisible by 9 because the sum of the digits is (22*45) which is divisible by 9. Subtracting 9 from the integer we get 1234 ... 4335, which is also divisible by 5, making it also divisible by 45. Thus the remainder is 9.

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 10 Problems and Solutions

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2017 AMC 12B Problems/Problem 19

Contents

Problem

Solution

Solution 2

See Also