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Difference between revisions of "2017 AMC 12B Problems/Problem 19"

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==Solution 2==
 
==Solution 2==
  
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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.
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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.
  
 
==See Also==
 
==See Also==

Revision as of 14:56, 4 February 2019

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.

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 12 Problems and Solutions
2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 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

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