Difference between revisions of "2020 AMC 10A Problems/Problem 24"
(→Solution) |
(→Solution) |
||
Line 8: | Line 8: | ||
We know that <math>gcd(63, n+120)=21</math>, so we can write <math>n+120\equiv0(mod 21)</math>. Simplifying, we get <math>n\equiv6(mod 21)</math>. Similarly, we can write <math>n+63\equiv0(mod60)</math>, or <math>n\equiv-3(mod60)</math>. Solving these two modular congruences, <math>n\equiv237(mod 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(mod 63)</math> so we try <math>n=1077+420=1497</math>. | We know that <math>gcd(63, n+120)=21</math>, so we can write <math>n+120\equiv0(mod 21)</math>. Simplifying, we get <math>n\equiv6(mod 21)</math>. Similarly, we can write <math>n+63\equiv0(mod60)</math>, or <math>n\equiv-3(mod60)</math>. Solving these two modular congruences, <math>n\equiv237(mod 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(mod 63)</math> so we try <math>n=1077+420=1497</math>. | ||
− | <math>1497+63\equiv0(mod 120) so again we add | + | <math>1497+63\equiv0(mod 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 conditions, so our answer is <math>1+9+1+7=\boxed{\textsf{(C) } 18}</math>. |
== Video Solution == | == Video Solution == |
Revision as of 02:25, 1 February 2020
Contents
Problem
Let be the least positive integer greater than for whichWhat is the sum of the digits of ?
Solution
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 conditions, so our 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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.