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

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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>
 
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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==See Also==
 
{{AMC8 box|year=2018|num-b=6|num-a=8}}
 
{{AMC8 box|year=2018|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 15:56, 21 November 2018

Problem 7

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 $\boxed{\textbf{(B) }3}$

See Also

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

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