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

Revision as of 00:39, 2 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 $1+9+1+7=\boxed{\textbf{(C) } 18}$ .

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 $\boxed{\textbf{(C) } 18}$ .

-Midnight

Solution 3 (bashing but worse)

Assume that $n$ has 4 digits. Then $n = abcd$ , where $a$ , $b$ , $c$ , $d$ represent digits of the number (not to get confused with $a * b * c * d$ ). As given the problem, $gcd(63, n + 120) = 21$ and $gcd(n + 63, 120) = 60$ . So we know that $d = 7$ (last digit of $n$ ). That means that $a + b + c + 7$

Video Solution

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

@@ Line 18: / Line 18: @@
 ==Solution 3 (bashing but worse)==
-Assume that <math>n</math> has 4 digits. Then <math>n = abcd</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> represent digits of the number (not to get confused with <math>a * b * c * d</math>).
+Assume that <math>n</math> has 4 digits. Then <math>n = abcd</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> represent digits of the number (not to get confused with <math>a * b * c * d</math>). As given the problem, <math>gcd(63, n + 120) = 21</math> and <math>gcd(n + 63, 120) = 60</math>. So we know that <math>d = 7</math> (last digit of <math>n</math>). That means that <math>a + b + c + 7</math>
 == Video Solution ==

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