2020 AMC 10A Problems/Problem 24
Contents
Problem
Let be the least positive integer greater than for whichWhat is the sum of the digits of ?
Solution 1
We know that , so we can write . Simplifying, we get . Similarly, we can write , or . Solving these two modular congruences, which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than , we find the least solution is . However, we are have not considered cases where or . so we try . so again we add to . It turns out that does indeed satisfy the original conditions, so our answer is .
Solution 2 (bashing)
We are given that and . This tells us that is divisible by but not . It also tells us that is divisible by 60 but not 120. Starting, we find the least value of which is divisible by which satisfies the conditions for , which is , making . We then now keep on adding until we get a number which satisfies the second equation. This number turns out to be , whose digits add up to .
-Midnight
Solution 3 (bashing but worse)
Assume that has 4 digits. Then , where , , , represent digits of the number (not to get confused with ). As given the problem, and . So we know that (last digit of ). That means that and . We can bash this after this. We just want to find all pairs of numbers such that is a multiple of 7 that is greater than a multiple of . Our equation for would be and our equation for would be , where is any integer. We plug this value in until we get a value of that makes satisfy the original problem statement (remember, ). After bashing for hopefully a couple minutes, we find that works. So which means that the sum of its digits is .
~ Baolan
Solution 4
The conditions of the problem reduce to the following. where and where . From these equations, we see that . Solving this diophantine equation gives us that , form. Since, is greater than , we can do some bounding and get that and . Now we start the bash by plugging in numbers that satisfy these conditions. We get , . So the answer is .
Video Solution
https://youtu.be/tk3yOGG2K-s -
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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