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

(Solution)
(Solution)
Line 5: Line 5:
 
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math>
 
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math>
  
== Solution ==
+
== 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{\textsf{(C) } 18}</math>.
 
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{\textsf{(C) } 18}</math>.
 +
 +
==Solution 2 (bashing)==
 +
 +
We are given that <math>gcd(63, n+120)=21</math> and
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 09:54, 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 1

We know that $gcd(63, n+120)=21$, so we can write $n+120\equiv0\pmod {21}$. Simplifying, we get $n\equiv6\pmod {21}$. Similarly, we can write $n+63\equiv0\pmod {60}$, or $n\equiv-3\pmod {60}$. Solving these two modular congruences, $n\equiv237\pmod {420}$ 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 $gcd(n+63, 120) =120$. ${1077+120}\equiv0\pmod {63}$ so we try $n=1077+420=1497$. ${1497+63}\equiv0\pmod {120}$ so again we add $420$ to $n$. It turns out that $n=1497+420=1917$ does indeed satisfy the original conditions, so our answer is $1+9+1+7=\boxed{\textsf{(C) } 18}$.

Solution 2 (bashing)

We are given that $gcd(63, n+120)=21$ and

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS