2016 AMC 8 Problems/Problem 5
Contents
Problem
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 1
From the second bullet point, we know that the second digit must be , for a number divisible by ends in zero. Since there is a remainder of when is divided by , the multiple of must end in a 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, .
~CHECKMATE2021
Solution 3
We know that the number has to be one more than a multiple of , because of the remainder of one, and the number has to be more than a multiple of , which means that it has to end in a . Now, if we just list the first few multiples of adding one to the number we get: . As we can see from these numbers, the only one that has a three in the denominator is , thus we divide by , getting , hence, . -fn106068
We could also remember that, for a two-digit number to be divisible by , the sum of its digits has to be equal to . Since the number is one more than a multiple of , the multiple we are looking for has a ones digit of , and therefore a tens digit of , and then we could proceed as above. -vaisri
Video Solution
https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3
A solution so simple a 12-year-old made it!
~Elijahman~
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=574
Video Solution
~savannahsolver
See Also
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.