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>. | ||
| − | + | -InterstellerApex | |
== Video Solution (CRITICAL THINKING!!!)== | == Video Solution (CRITICAL THINKING!!!)== | ||
Revision as of 20:47, 9 January 2024
Contents
Problem
The 
-digit number 
is divisible by 
. What is the remainder when this number is divided by 
?
Solution
We use the property that the digits of a number must sum to a multiple of if it are divisible by . This means 
must be divisible by . The only possible value for 
then must be 
. Since we are looking for the remainder when divided by , we can ignore the thousands. The remainder when 
is divided by is 
.
-InterstellerApex
Video Solution (CRITICAL THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=2341
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
