Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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− | We are given that <math>\gcd(63, n+120)=21</math> and <math>\gcd(n+63,120) = 60</math>. This tells us that <math>n+120</math> is divisible by <math>21</math> but not <math>63</math>. It also tells us that <math>n+63</math> is divisible by 60 but not 120. Starting, we find the least value of <math>n+120</math> which is divisible by <math>21</math> which satisfies the conditions for <math>n</math>, which is <math>1134</math>, making <math>n | + | We are given that <math>\gcd(63, n+120)=21</math> and <math>\gcd(n+63,120) = 60</math>. This tells us that <math>n+120</math> is divisible by <math>21</math> but not <math>63</math>. It also tells us that <math>n+63</math> is divisible by 60 but not 120. Starting, we find the least value of <math>n+120</math> which is divisible by <math>21</math> which satisfies the conditions for <math>n</math>, which is <math>1134</math>, making <math>n=1014</math>. We then now keep on adding <math>21</math> until we get a number which satisfies the second equation. This number turns out to be <math>1917</math>, whose digits add up to <math>\boxed{\text{C}, 18}</math>. |
+ | |||
+ | -Midnight | ||
== Video Solution == | == Video Solution == |
Revision as of 09:02, 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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