2018 AMC 8 Problems/Problem 21
Contents
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
Solution 4
We can use modular arithmetic. Set up the equations: and These equations can also be written as and Since is congruent to numbers and then it must also be congruent to their LCM. Thus, since 198 is the LCM of and Since these numbers have to be three digits, they can only be and This gives us the answer of
~ethancui0529
Solution 5
Let be the three digit positive integer. . Then, we add four to all sides and write . Now, we know that is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so . From this, we can figure out that can be 5 different three digit numbers -- and . , the answer is
~DY
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=939
~ pi_is_3.14
Video Solutions
- Happytwin
~savannahsolver
https://www.youtube.com/watch?v=PjYwbGm_2aM ~David
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.