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Difference between revisions of "2017 AMC 10B Problems/Problem 23"

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We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
 
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
 
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
 
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
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The remainder when divided by 9 is <math>1+2+3+4 \cdot 1+0+1+1 \cdot 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 \cdot 10+11+12 \cdot 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which clearly has a remainder of 0 mod 9. Therefore, by CRT, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
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The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4 \cdot 1+0+1+1 \cdot 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 \cdot 10+11+12 \cdot 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Therefore, by CRT, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
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Note: the sum of the digits of <math>N</math> is <math>270</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:44, 16 February 2017

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 $1+2+3+4 \cdot 1+0+1+1 \cdot 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 \cdot 10+11+12 \cdot 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$, 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 $270$

See Also

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