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

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 $n+63\equiv0(\mod 60)$, or $n\equiv-3(\mod 60)$. 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 $420$ to $n$. It turns out that $n=1497+420=1917$ does indeed satisfy the original conditions, so our answer is $1+9+1+7=\boxed{\textsf{(C) } 18}$.

Video Solution

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