2020 AMC 10A Problems/Problem 24

Revision as of 10:02, 1 February 2020 by Xmidnightfirex (talk | contribs) (Solution 2 (bashing))

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 $\gcd(n+63,120) = 60$. This tells us that $n+120$ is divisible by $21$ but not $63$. It also tells us that $n+63$ is divisible by 60 but not 120. Starting, we find the least value of $n+120$ which is divisible by $21$ which satisfies the conditions for $n$, which is $1134$, making $n$ $1014$. We then now keep on adding $21$ until we get a number which satisfies the second equation. This number turns out to be $1917$, whose digits add up to $\boxed{\text{C}, 18}$.

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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