2021 Fall AMC 10B Problems/Problem 22
Contents
Problem
For each integer , let be the sum of all products , where and are integers and . What is the sum of the 10 least values of such that is divisible by ?
Solution 1
To get from to , we add .
Now, we can look at the different values of mod . For and , then we have . However, for , we have
Clearly, Using the above result, we have , and , , and are all divisible by . After , we have , , and all divisible by , as well as , and . Thus, our answer is -BorealBear
-minor edit by Yiyj1
Solution 2 (bash)
Since we have a wonky function, we start by trying out some small cases and see what happens. If is and is , then there is one case. We have mod for this case. If is , we have which is still mod . If is , we have to add which is a multiple of , meaning that we are still at mod . If we try a few more cases, we find that when is , we get a multiple of . When is , we are adding mod , and therefore, we are still at a multiple of .
When is , then we get mod + which is times a multiple of . Therefore, we have another multiple of . When is , so we have mod . So, every time we have mod , mod , and mod , we always have a multiple of . Think about it: When is , it will have to be , so it is a multiple of . Therefore, our numbers are . Adding the numbers up, we get
~Arcticturn
Solution 3
Denote , and .
Hence, .
Therefore,
Hence, is divisible by 3 if and only if is divisible by .
First, is always divisible by 8. Otherwise, is not even an integer.
Second, we find conditions for , such that is divisible by 9.
Because is not divisible by 3, it cannot be divisible by 9.
Hence, we need to find conditions for , such that is divisible by 9. This holds of .
Therefore, the 10 least values of such that is divisible by 9 (equivalently, is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35. Their sum is 197.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
~Interstigation
Video Solution 2 by WhyMath
~savannahsolver
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AMC 10 Problems and Solutions |
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