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

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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 + 57</math>, <math>l = 7a + 19</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 = 53</math>, <math>k = 97</math>. So the answer is <math>\boxed{1917}</math>.
 
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 + 57</math>, <math>l = 7a + 19</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 = 53</math>, <math>k = 97</math>. So the answer is <math>\boxed{1917}</math>.
 
==Solution 5==
 
==Solution 5==
−
<math> You can first find that n must be congruent to </math>6 mod 21<math> and </math>57 mod 60<math>. The we can find that </math>n=21x+6<math> and </math>n=60y+57<math>, 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 </math>1+9+1+7<math>=</math>\boxed{1917}<math>.</math>
+
You can first find that n must be congruent to 6 mod 21 and 57 mod 60. The we can find that <math>n=21x+6</math> and <math>n=60y+57</math>, 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 <math>1+9+1+7</math>=<math>\boxed{1917}</math>.
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 23:45, 3 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 $12 + abc \equiv0\pmod {7}$ 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 $k = 20a + 57$, $l = 7a + 19$ 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 $l = 53$, $k = 97$. So the answer is $\boxed{1917}$.

Solution 5

You can first find that n must be congruent to 6 mod 21 and 57 mod 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 $1+9+1+7$=$\boxed{1917}$.

Video Solution

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

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
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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