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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, thus 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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