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Difference between revisions of "2018 AMC 8 Problems/Problem 7"

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==Solution==
 
==Solution==
  
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We use the property that the digits of a number must sum to a multiple of <math>9</math> if it are divisible by <math>9</math>. This means <math>2+0+1+8+U</math> must be divisible by <math>9</math>. The only possible value for <math>U</math> then must be <math>7</math>. Since we are looking for the remainder when divided by <math>8</math>, we can ignore the thousands. The remainder when <math>187</math> is divided by <math>8</math> is <math>\boxed{\textbf{(B) }3}</math>
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We use the property that the digits of a number must sum to a multiple of <math>9</math> if it are divisible by <math>9</math>. This means <math>2+0+1+8+U</math> must be divisible by <math>9</math>. The only possible value for <math>U</math> then must be <math>7</math>. Since we are looking for the remainder when divided by <math>8</math>, we can ignore the thousands. The remainder when <math>187</math> is divided by <math>8</math> is <math>\boxed{\textbf{(B) }3}</math>.
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20187 mod 8 =_ 3
 
20187 mod 8 =_ 3

Revision as of 06:47, 26 December 2022

Problem

The $5$-digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution

We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for $U$ then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $\boxed{\textbf{(B) }3}$.

20187 mod 8 =_ 3

Video Solution

https://youtu.be/6xNkyDgIhEE?t=2341

https://youtu.be/doHZiAT36BY

~savannahsolver

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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