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

Revision as of 09:02, 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 $\gcd(n+63,120) = 60$ . 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 $\boxed{\text{C}, 18}$ .

-Midnight

Video Solution

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

@@ Line 11: / Line 11: @@
 ==Solution 2 (bashing)==
-We are given that <math>\gcd(63, n+120)=21</math> and <math>\gcd(n+63,120) = 60</math>. This tells us that <math>n+120</math> is divisible by <math>21</math> but not <math>63</math>. It also tells us that <math>n+63</math> is divisible by 60 but not 120. Starting, we find the least value of <math>n+120</math> which is divisible by <math>21</math> which satisfies the conditions for <math>n</math>, which is <math>1134</math>, making <math>n</math> <math>1014</math>. We then now keep on adding <math>21</math> until we get a number which satisfies the second equation. This number turns out to be <math>1917</math>, whose digits add up to <math>\boxed{\text{C}, 18}</math>.
+We are given that <math>\gcd(63, n+120)=21</math> and <math>\gcd(n+63,120) = 60</math>. This tells us that <math>n+120</math> is divisible by <math>21</math> but not <math>63</math>. It also tells us that <math>n+63</math> is divisible by 60 but not 120. Starting, we find the least value of <math>n+120</math> which is divisible by <math>21</math> which satisfies the conditions for <math>n</math>, which is <math>1134</math>, making <math>n=1014</math>. We then now keep on adding <math>21</math> until we get a number which satisfies the second equation. This number turns out to be <math>1917</math>, whose digits add up to <math>\boxed{\text{C}, 18}</math>.
+-Midnight
 == Video Solution ==

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