Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math> | ||
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From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one: | From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one: | ||
Revision as of 13:27, 5 January 2017
The number 
is a two-digit number.
• When is divided by 
, the remainder is 
.
• When is divided by 
, the remainder is 
.
What is the remainder when is divided by 
?
Solution
From the second bullet point, we know that the second digit must be . Because there is a remainder of when it is divided by , the multiple of must end in a 
. We now look for this one:
The number 
satisfies both conditions. We subtract the biggest multiple of less than 
to get the remainder. Thus, 
.
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
