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 $100 ≤ 11z+7 <1000$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)), it gives us $9 ≤ z ≤90$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)), 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 $18(LCM of 6 & 9)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.). So let's say , then $9 ≤ z = 18p-1 ≤ 90, 1 ≤ p ≤ 91/18 or 1 ≤ p ≤ 5$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)). Thus our answer is ~LarryFlora
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 | |
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