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

Because we know that $gcd(63, n+120)=21$ , 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(mod63) $so we try$ n=1077+420=1497 $.$ 1497+63\equiv0(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 - $Phineas1500$

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2020 AMC 10A Problems/Problem 24

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