2018 AMC 8 Problems/Problem 21
Problem
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
Solution 1
Looking at the values, we notice that , and . This means we are looking for a value that is four less than a multiple of , , and . The least common multiple of these numbers is , so the numbers that fulfill this can be written as , where is a positive integer. This value is only a three digit integer when is or , which gives and respectively. Thus we have values, so our answer is
Solution 2
Let us create the equations: , and we know , it gives us , which is the range of the value of z. Because of , then , so must be a mutiple of 6. Because of , then , so must also be a mutiple of . Hence, the value of must be a common multiple of and , which means multiples of . So let's say , then , so . Thus the answer is ~LarryFlora
Solution 3
By the Chinese Remainder Theorem, we have that all solutions are in the form where Counting the number of values, we get
~mathboy282
Video Solutions
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14
~savannahsolver
https://www.youtube.com/watch?v=PjYwbGm_2aM
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.