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

m (→‎Video Solution)
(Added a Solution + Remark)
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-SmileKat32
 
-SmileKat32
  
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==Solution 8==
 +
We are able to set-up the following system-of-congruences:
 +
<cmath>n \equiv 6 \pmod {21},</cmath>
 +
<cmath>n \equiv 57 \pmod {60}.</cmath>
 +
Therefore, by definition, we are able to set-up the following system of equations:
 +
<cmath>n = 21a + 6,</cmath>
 +
<cmath>n = 60b + 57.</cmath>
 +
Thus,
 +
<cmath>21a + 6 = 60b + 57</cmath>
 +
<cmath>\implies 7a + 2 = 20b + 19.</cmath>
 +
We know <math>7a \equiv 0 \pmod {7},</math> and since <math>7a = 20b + 17,</math> therefore <math>20b + 17 \equiv 0 \pmod{7}.</math> Simplifying this congruence further, we have
 +
<cmath>5b \equiv 1 \pmod{7}</cmath>
 +
<cmath>\implies b \equiv 3 \pmod {7}.</cmath>
 +
Thus, by definition, <math>b = 7x + 3.</math> Substituting this back into our original equation,
 +
<cmath>n = 60(7x + 3) + 57</cmath>
 +
<cmath>\implies n = 420x + 180 + 57</cmath>
 +
<cmath>\implies n = 420x + 237.</cmath>
 +
By definition, we are able to set-up the following congruence:
 +
<cmath>n \equiv 237 \pmod{420}.</cmath>
 +
Thus, <math>n = 1617</math>, so our answer is simply <math>\boxed{18}</math>.
 +
 +
(Remarks. <math>n \equiv 6 \pmod{21}</math> since <math>n \equiv -120 \pmod{21},</math> by definition & <math>n \equiv 57 \pmod{60}</math> since <math>n \equiv -63 \pmod{60},</math> by definition.
 +
 +
Remember,  <math>5b \equiv 1 \pmod{7} \implies 5b \equiv 15 \pmod{7} \implies b \equiv 3 \pmod{7}.</math>
 +
 +
Lastly, the reason why <math>n \neq 1077</math> is <math>n + 120</math> would be divisible by <math>63</math>, which is not possible due to the certain condition.)
 +
 +
~ nikenissan
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 17:21, 7 August 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

Solution 8

We are able to set-up the following system-of-congruences: \[n \equiv 6 \pmod {21},\] \[n \equiv 57 \pmod {60}.\] Therefore, by definition, we are able to set-up the following system of equations: \[n = 21a + 6,\] \[n = 60b + 57.\] Thus, \[21a + 6 = 60b + 57\] \[\implies 7a + 2 = 20b + 19.\] We know $7a \equiv 0 \pmod {7},$ and since $7a = 20b + 17,$ therefore $20b + 17 \equiv 0 \pmod{7}.$ Simplifying this congruence further, we have \[5b \equiv 1 \pmod{7}\] \[\implies b \equiv 3 \pmod {7}.\] Thus, by definition, $b = 7x + 3.$ Substituting this back into our original equation, \[n = 60(7x + 3) + 57\] \[\implies n = 420x + 180 + 57\] \[\implies n = 420x + 237.\] By definition, we are able to set-up the following congruence: \[n \equiv 237 \pmod{420}.\] Thus, $n = 1617$, so our answer is simply $\boxed{18}$.

(Remarks. $n \equiv 6 \pmod{21}$ since $n \equiv -120 \pmod{21},$ by definition & $n \equiv 57 \pmod{60}$ since $n \equiv -63 \pmod{60},$ by definition.

Remember, $5b \equiv 1 \pmod{7} \implies 5b \equiv 15 \pmod{7} \implies b \equiv 3 \pmod{7}.$

Lastly, the reason why $n \neq 1077$ is $n + 120$ would be divisible by $63$, which is not possible due to the certain condition.)

~ nikenissan

Video Solution

https://youtu.be/tk3yOGG2K-s

Video Solution

https://www.youtube.com/watch?v=gdGmSyzR908&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=5 ~ MathEx

See Also

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

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