Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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== Solution 1== | == Solution 1== | ||
− | We know that <math>gcd(63, n+120)=21</math>, so we can write <math>n+120\equiv0\pmod {21}</math>. Simplifying, we get <math>n\equiv6\pmod {21}</math>. Similarly, we can write <math>n+63\equiv0\pmod {60}</math>, or <math>n\equiv-3\pmod {60}</math>. Solving these two modular congruences, <math>n\equiv237\pmod {420}</math> 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 <math>1000</math>, we find the least solution is <math>n=1077</math>. However, we are have not considered cases where <math>gcd(63, n+120) =63</math> or <math>gcd(n+63, 120) =120</math>. <math>{1077+120}\equiv0\pmod {63}</math> so we try <math>n=1077+420=1497</math>. <math>{1497+63}\equiv0\pmod {120}</math> so again we add <math>420</math> to <math>n</math>. It turns out that <math>n=1497+420=1917</math> does indeed satisfy the original conditions, so our answer is <math>1+9+1+7=\boxed{\ | + | We know that <math>gcd(63, n+120)=21</math>, so we can write <math>n+120\equiv0\pmod {21}</math>. Simplifying, we get <math>n\equiv6\pmod {21}</math>. Similarly, we can write <math>n+63\equiv0\pmod {60}</math>, or <math>n\equiv-3\pmod {60}</math>. Solving these two modular congruences, <math>n\equiv237\pmod {420}</math> 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 <math>1000</math>, we find the least solution is <math>n=1077</math>. However, we are have not considered cases where <math>gcd(63, n+120) =63</math> or <math>gcd(n+63, 120) =120</math>. <math>{1077+120}\equiv0\pmod {63}</math> so we try <math>n=1077+420=1497</math>. <math>{1497+63}\equiv0\pmod {120}</math> so again we add <math>420</math> to <math>n</math>. It turns out that <math>n=1497+420=1917</math> does indeed satisfy the original conditions, so our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. |
==Solution 2 (bashing)== | ==Solution 2 (bashing)== |
Revision as of 19:41, 1 February 2020
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
Video Solution
https://youtu.be/tk3yOGG2K-s -
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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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