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

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We know that this number is divisible by <math>9</math> because the sum of the digits is <math>270</math>, which is divisible by <math>9</math>. If we subtracted <math>9</math> from the integer we would get <math>1234 \cdots 4335</math>, which is also divisible by <math>5</math> and by <math>45</math>. Thus the remainder is <math>9</math>, or <math>\boxed{\textbf{B}}</math>.
 
We know that this number is divisible by <math>9</math> because the sum of the digits is <math>270</math>, which is divisible by <math>9</math>. If we subtracted <math>9</math> from the integer we would get <math>1234 \cdots 4335</math>, which is also divisible by <math>5</math> and by <math>45</math>. Thus the remainder is <math>9</math>, or <math>\boxed{\textbf{B}}</math>.
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 +
==Solution 3 (Beginner's Method)==
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 +
To find the sum of digits of our number, we break it up into <math>5</math> cases, starting with <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>.
 +
 +
Case 1: <math>1+2+3+\cdots+9 = 45</math>
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Case 2: <math>1+0+1+1+1+2+\cdots+1+8+1+9 = 55</math> (We add 10 to the previous cases, as we are in the next ten's place)
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Case 3: <math>2+0+2+1+\cdots+2+9 = 65</math>
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Case 4: <math>3+0+3+1+\cdots+3+9 = 75</math>
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Case 5: <math>4+0+4+1+\cdots+4+4 = 30</math>
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Thus the sum of the digits is <math>45+55+65+75+30 = 270</math>, so the number is divisible by <math>9</math>. We notice that the number ends in "<math>4</math>", which is <math>9</math> more than a multiple of <math>5</math>. Thus if we subtracted <math>9</math> from our number it would be divisible by <math>5</math>, and <math>5\cdot 9 = 45</math>. (Multiple of n - n = Multiple of n)
 +
 +
So our remainder is <math>\boxed{textbf{(C)}9</math>, the value we need to add to the multiple of 45 to get to our number.
  
 
==See Also==
 
==See Also==

Revision as of 15:15, 11 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 $\boxed {\textbf {(C)}}$.

Solution 2

We know that this number is divisible by $9$ because the sum of the digits is $270$, which is divisible by $9$. If we subtracted $9$ from the integer we would get $1234 \cdots 4335$, which is also divisible by $5$ and by $45$. Thus the remainder is $9$, or $\boxed{\textbf{B}}$.

Solution 3 (Beginner's Method)

To find the sum of digits of our number, we break it up into $5$ cases, starting with $0$, $1$, $2$, $3$, or $4$.

Case 1: $1+2+3+\cdots+9 = 45$ Case 2: $1+0+1+1+1+2+\cdots+1+8+1+9 = 55$ (We add 10 to the previous cases, as we are in the next ten's place) Case 3: $2+0+2+1+\cdots+2+9 = 65$ Case 4: $3+0+3+1+\cdots+3+9 = 75$ Case 5: $4+0+4+1+\cdots+4+4 = 30$

Thus the sum of the digits is $45+55+65+75+30 = 270$, so the number is divisible by $9$. We notice that the number ends in "$4$", which is $9$ more than a multiple of $5$. Thus if we subtracted $9$ from our number it would be divisible by $5$, and $5\cdot 9 = 45$. (Multiple of n - n = Multiple of n)

So our remainder is $\boxed{textbf{(C)}9$ (Error compiling LaTeX. Unknown error_msg), the value we need to add to the multiple of 45 to get to our number.

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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