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 (z+1) must be mutiples of 6. Because of 
, then 
, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples 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 Solution
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
