Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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==Solution== | ==Solution== | ||
From the second bullet, we know that the second digit must be 3. Because there is a remainder of 1 when it is divided by three, the multiple of 9 must end in a 2. We now look for this one: | From the second bullet, we know that the second digit must be 3. Because there is a remainder of 1 when it is divided by three, the multiple of 9 must end in a 2. We now look for this one: | ||
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<math>9(1)=9\\ | <math>9(1)=9\\ | ||
9(2)=18\\ | 9(2)=18\\ | ||
Revision as of 09:13, 23 November 2016
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, we know that the second digit must be 3. Because there is a remainder of 1 when it is divided by three, the multiple of 9 must end in a 2. We now look for this one:
The number 
satisfies both conditions. We subtract the biggest multiple of 11 less than 73 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. 
