Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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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> in order for it to have the desired remainder<math>\pmod {10}.</math> | + | 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> in order for it to have the desired remainder<math>\pmod {10}.</math> We now look for this one: |
<math>9(1)=9\\ | <math>9(1)=9\\ | ||
Revision as of 22:13, 1 November 2020
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 
?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=574
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 
in order for it to have the desired remainder
We now look for this one:
The number 
satisfies both conditions. We subtract the biggest multiple of less than 
to get the remainder. Thus, 
.
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
| 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. 
