Difference between revisions of "2020 AMC 10A Problems/Problem 24"

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== Solution ==
 
== Solution ==
  
Because we know that <math>gcd(63, n+120)=21</math>, 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 or </math>gcd(n+63, 120) =120<math>. </math>1077+120\equiv0(mod63)<math> so we try </math>n=1077+420=1497<math>. </math>1497+63\equiv0(120) so again we add another <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>.
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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 another </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}$.
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 02:25, 1 February 2020

Problem

Let $n$ be the least positive integer greater than $1000$ for which\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]What is the sum of the digits of $n$?

$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution

We know that $gcd(63, n+120)=21$, so we can write $n+120\equiv0(mod 21)$. Simplifying, we get $n\equiv6(mod 21)$. Similarly, we can write $n+63\equiv0(mod60)$, or $n\equiv-3(mod60)$. Solving these two modular congruences, $n\equiv237(mod 420)$ 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 $1000$, we find the least solution is $n=1077$. However, we are have not considered cases where $gcd(63, n+120) =63$ or $gcd(n+63, 120) =120$. $1077+120\equiv0(mod 63)$ so we try $n=1077+420=1497$.

$1497+63\equiv0(mod 120) so again we add another$420$to$n$. It turns out that$n=1497+420=1917$does indeed satisfy the conditions, so our answer is$1+9+1+7=\boxed{\textsf{(C) } 18}$.

Video Solution

https://youtu.be/tk3yOGG2K-s - $Phineas1500$

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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