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

Revision as of 21:56, 13 April 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 used to so,be a system of MODULAR CONGURENCES). 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 $12 + abc \equiv0\pmod {7}$ and $7 + abc\equiv0\pmod {6}$ . We can bash this after this. We just want to find all pairs of numbers $(x, y)$ such that $x$ is a multiple of 7 that is $5$ greater than a multiple of $6$ . Our equation for $12 + abc$ would be $42*j + 35 = x$ and our equation for $7 + abc$ would be $42*j + 30 = y$ , where $j$ is any integer. We plug this value in until we get a value of $abc$ that makes $n = abc7$ satisfy the original problem statement (remember, $abc > 100$ ). After bashing for hopefully a couple minutes, we find that $abc = 191$ works. So $n = 1917$ which means that the sum of its digits is $\boxed{\textbf{(C) } 18}$ .

~ Baolan

Solution 4

The conditions of the problem reduce to the following. $n+120 = 21k$ where $gcd(k,3) = 1$ and $n+63 = 60l$ where $gcd(l,2) = 1$ . From these equations, we see that $21k - 60l = 57$ . Solving this diophantine equation gives us that $k = 20a + 17$ , $l = 7a + 5$ form. Since, $n$ is greater than $1000$ , we can do some bounding and get that $k > 53$ and $l > 17$ . Now we start the bash by plugging in numbers that satisfy these conditions. We get $l = 33$ , $k = 97$ . So the answer is $1917 \implies 1+9+1+7= \boxed{\textbf{(C) } 18}$ .

Edited by ~fastnfurious1

Solution 5

You can first find that n must be congruent to $6\equiv0\pmod {21}$ and $57\equiv0\pmod {60}$ . The we can find that $n=21x+6$ and $n=60y+57$ , where x and y are integers. Then we can find that y must be odd, since if it was even the gcd will be 120, not 60. Also, the unit digit of n has to be 7, since the unit digit of 60y is always 0 and the unit digit of 57 is 7. Therefore, you can find that x must end in 1 to satisfy n having a unit digit of 7. Also, you can find that x must not be a multiple of three or else the gcd will be 63. Therefore, you can test values for x and you can find that x=91 satisfies all these conditions.Therefore, n is 1917 and $1+9+1+7$ = $\boxed{(C)18}$ .-happykeeper

Solution 6 (Reverse Euclidean Algorithm)

We are given that $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60.$ By applying the Euclidean algorithm, but in reverse, we have $\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21$ and $\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.$

We now know that $n+183$ must be divisible by $21$ and $60,$ so it is divisible by $\text{lcm}(21, 60) = 420.$ Therefore, $n+183 = 420k$ for some integer $k.$ We know that $3 \nmid k,$ or else the first condition won't hold ( $\gcd$ will be $63$ ) and $2 \nmid k,$ or else the second condition won't hold ( $\gcd$ will be $120$ ). Since $k = 1$ gives us too small of an answer, then $k=5 \implies n = 1917,$ so the answer is $1+9+1+7 = \boxed{(C)18}.$

Solution 7

$\gcd(n+63,120)=60$ tells us $n+63\equiv60\pmod {120}$ . The smallest $n+63$ that satisfies the previous condition and $n>1000$ is $1140$ , so we start from there. If $n+63=1140$ , then $n+120=1197$ . Because $\gcd(n+120,63)=21$ , $n+120\equiv21\pmod {63}$ or $n+120\equiv42\pmod {63}$ . We see that $1197\equiv0\pmod {63}$ , which does not fulfill the requirement for $n+120$ , so we continue by keep on adding $120$ to $1197$ , in order to also fulfill the requirement for $n+63$ . Soon, we see that $n+120\pmod {63}$ decreases by $6$ every time we add $120$ , so we can quickly see that $n=1917$ because at that point $n+120\equiv21\pmod {63}$ . Adding up all the digits in $1917$ , we have $\boxed{(C)18}$ .

-SmileKat32

Video Solution

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

@@ Line 22: / Line 22: @@
 ==Solution 4==
-The conditions of the problem reduce to the following. <math>n+120 = 21k</math> where <math>gcd(k,3) = 1</math> and <math>n+63 = 60l</math> where <math>gcd(l,2) = 1</math>. From these equations, we see that <math>21k - 60l = 57</math>. Solving this diophantine equation gives us that <math>k = 20a + 19</math>, <math>l = 7a + 5</math> form. Since, <math>n</math> is greater than <math>1000</math>, we can do some bounding and get that <math>k > 53</math> and <math>l > 17</math>. Now we start the bash by plugging in numbers that satisfy these conditions. We get <math>l = 33</math>, <math>k = 97</math>. So the answer is <math>1917 \implies 1+9+1+7=
+The conditions of the problem reduce to the following. <math>n+120 = 21k</math> where <math>gcd(k,3) = 1</math> and <math>n+63 = 60l</math> where <math>gcd(l,2) = 1</math>. From these equations, we see that <math>21k - 60l = 57</math>. Solving this diophantine equation gives us that <math>k = 20a + 17</math>, <math>l = 7a + 5</math> form. Since, <math>n</math> is greater than <math>1000</math>, we can do some bounding and get that <math>k > 53</math> and <math>l > 17</math>. Now we start the bash by plugging in numbers that satisfy these conditions. We get <math>l = 33</math>, <math>k = 97</math>. So the answer is <math>1917 \implies 1+9+1+7=
 \boxed{\textbf{(C) } 18}</math>.

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