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

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== Solution 1 ==  
 
== Solution 1 ==  
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We know that <math>\gcd(n+57,63)=21</math> and <math>\gcd(n-57, 120)= 60</math> by the Euclidean Algorithm. Hence, let <math>n+57=21\alpha</math> and <math>n-57=60 \gamma</math>, where <math>\gcd(\alpha,3)=1</math> and <math>\gcd(\gamma,2)=1</math>. Subtracting the two equations, <math>38=7\alpha-20\gamma</math>. Letting <math>\gamma = 2s+1</math>, we get <math>58=7\alpha-40s</math>. Taking modulo <math>40</math>, we have <math>\alpha \equiv{14} \pmod{40}</math>. We are given that <math>n=21\alpha -57 >1000</math>, so <math>\alpha \geq 51</math>. Notice that if <math>\alpha =54</math> then the condition <math>\gcd(\alpha,3)=1</math> is violated. The next possible value of  <math>\alpha = 94</math> satisfies the given condition, giving us the answer <math>1917</math>.
+
We know that <math>\gcd(n+57,63)=21</math> and <math>\gcd(n-57, 120)= 60</math> by the Euclidean Algorithm. Hence, let <math>n+57=21\alpha</math> and <math>n-57=60 \gamma</math>, where <math>\gcd(\alpha,3)=1</math> and <math>\gcd(\gamma,2)=1</math>. Subtracting the two equations, <math>38=7\alpha-20\gamma</math>. Letting <math>\gamma = 2s+1</math>, we get <math>58=7\alpha-40s</math>. Taking modulo <math>40</math>, we have <math>\alpha \equiv{14} \pmod{40}</math>. We are given that <math>n=21\alpha -57 >1000</math>, so <math>\alpha \geq 51</math>. Notice that if <math>\alpha =54</math> then the condition <math>\gcd(\alpha,3)=1</math> is violated. The next possible value of  <math>\alpha = 94</math> satisfies the given condition, giving us the answer <math>n=1917</math>.
  
 
Alternatively, we could have said <math>\alpha = 40k+14 \equiv{0} \pmod{3}</math> for <math>k \equiv{1} \pmod{3}</math> only, so <math>k \equiv{0,2} \pmod{3}</math>, giving us our answer. Since the problem asks for the sum of the digits of <math>n</math>, our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>.
 
Alternatively, we could have said <math>\alpha = 40k+14 \equiv{0} \pmod{3}</math> for <math>k \equiv{1} \pmod{3}</math> only, so <math>k \equiv{0,2} \pmod{3}</math>, giving us our answer. Since the problem asks for the sum of the digits of <math>n</math>, our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>.

Revision as of 09:33, 14 January 2022

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 $\gcd(n+57,63)=21$ and $\gcd(n-57, 120)= 60$ by the Euclidean Algorithm. Hence, let $n+57=21\alpha$ and $n-57=60 \gamma$, where $\gcd(\alpha,3)=1$ and $\gcd(\gamma,2)=1$. Subtracting the two equations, $38=7\alpha-20\gamma$. Letting $\gamma = 2s+1$, we get $58=7\alpha-40s$. Taking modulo $40$, we have $\alpha \equiv{14} \pmod{40}$. We are given that $n=21\alpha -57 >1000$, so $\alpha \geq 51$. Notice that if $\alpha =54$ then the condition $\gcd(\alpha,3)=1$ is violated. The next possible value of $\alpha = 94$ satisfies the given condition, giving us the answer $n=1917$.

Alternatively, we could have said $\alpha = 40k+14 \equiv{0} \pmod{3}$ for $k \equiv{1} \pmod{3}$ only, so $k \equiv{0,2} \pmod{3}$, giving us our answer. Since the problem asks for the sum of the digits of $n$, our answer is $1+9+1+7=\boxed{\textbf{(C) } 18}$.

~Prabh1512, with edits by Terribleteeth.

Solution 2

The conditions of the problem reduce to the following: $n+120 = 21k$ and $n+63 = 60\ell$, where $\gcd(k,3) = 1$ and $\gcd(\ell,2) = 1$. From these equations, we see that $21k - 60\ell = 57$. Solving this Diophantine equation gives us that $k = 20a + 17$ and $\ell = 7a + 5$. Since, $n>1000$, we can do some bounding and get that $k > 53$ and $\ell > 17$. Now we start bashing by plugging in numbers that satisfy these conditions: $a=4$ is the first number that works so we get $\ell = 33$, $k = 97$. Therefore, we have $n=21(97)-120=60(33)-63=1917$, and our answer is $1+9+1+7=\boxed{\textbf{(C) } 18}$.

Solution 4

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 $1+9+1+7 = \boxed{\textbf{(C) } 18}$.-happykeeper

Solution 5 (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 $1+9+1+7 = \boxed{\textbf{(C) } 18}.$

Solution 6

$\gcd(n+63,120)=60$ 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 $n+120\equiv21\pmod {63}$. We add up all digits of $1917$: $\boxed{\textbf{(C) } 18}$.

~SmileKat32

Solution 7

We are able to set up the following system of congruences: \begin{align*} n &\equiv 6 \pmod {21}, \\ n &\equiv 57 \pmod {60}. \end{align*} Therefore, by definition, we are able to set-up the following system of equations: \begin{align*} n &= 21a + 6, \\ n &= 60b + 57. \end{align*} Thus, we have $21a + 6 = 60b + 57,$ from which \[7a + 2 = 20b + 19.\] 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 \[b\equiv 3 \pmod{7}.\] Thus, by definition, $b = 7x + 3.$ Substituting this back into our original equation, we get \[n = 60(7x + 3) + 57 = 420x + 237.\] By definition, we are able to set up the following congruence: \[n \equiv 237 \pmod{420}.\] Thus, $n = 1917$, so our answer is simply $1+9+1+7=\boxed{\textbf{(C) } 18}$.

Remark

Since $n \equiv -120 \pmod{21},$ we conclude that $n \equiv 6 \pmod{21}.$

Since $n \equiv -63 \pmod{60},$ we conclude that $n \equiv 57 \pmod{60}.$

Remember that $b \equiv 3 \pmod{7}.$

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 ~Midnight

Solution 8

First, we find $n$. We know that it is greater than $1000$, so we first input $n = 1000$. From the first equation, $\gcd(63, n + 120) = 21$, we know that if $n$ is correct, after we add $120$ to it, it should be divisible by $21$, but not $63$: \[\frac{n + 120}{21} = \frac{1120}{21} = 53\text{ R }7.\] This does not work. To get to the nearest number divisible by $21$, we have to add $14$ to cancel out the remainder. (Note that we don't subtract $7$ to get to $53$; $n$ is already at its lowest possible value!) Adding $14$ to $1000$ gives us $n = 1014$. (Note: $n$ is currently divisible by 63, but that's fine since we'll be changing it in the next step.)

Now using the second equation, $\gcd(n + 63, 120) = 60$, we know that if $n$ is correct, then $n+63$ is divisible by $60$ but not $120$: \[\frac{n + 63}{60} = \frac{1077}{60} = 17\text{ R }57.\] Again, this does not work. This requires some guessing and checking. We can add $21$ over and over again until $n$ is valid. This changes $n$ while also maintaining that $\frac{n + 120}{21}$ has no remainders. After adding $21$ once, we get $18 r 18$. By pure luck, adding $21$ two more times gives us $19$ with no remainders. We now have $1077 + 21 + 21 + 21 = 1140$. However, this number is divisible by $120$. To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$. Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$. Adding $420$ again gives us $1980$, which is valid. However, remember that this is equal to $n + 63$, so subtracting $63$ from $1980$ gives us $1917$, which is $n$.

The sum of its digits are $1 + 9 + 1 + 7 = \boxed{\textbf{(C) } 18}$.

~primegn

Solution 9 (Euclidean Algorithm)

By the Euclidean Algorithm, we have \begin{alignat*}{8} \gcd(63,n+120)&=\hspace{1mm}&&\gcd(63,\phantom{ }\underbrace{n+120-63k_1}_{(n+120) \ \mathrm{mod} \ 63}\phantom{ })&&=21, &&\hspace{10mm}(1) \\ \gcd(n+63,120)&=&&\gcd(\phantom{ }\underbrace{n+63-120k_2}_{(n+63) \ \mathrm{mod} \ 120}\phantom{ },120)&&=60.&&\hspace{10mm}(2) \end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$

More generally, let $t\in\{1,2\},$ so we get \begin{align*} n+120-63k_1&=21t, &\hspace{55.5mm}(1*) \\ n+63-120k_2&=60. &\hspace{55.5mm}(2*) \end{align*} Subtracting $(2*)$ from $(1*)$ and then simplifying give \begin{align*} 57-63k_1+120k_2&=21t-60 \\ 117-63k_1+120k_2&=21t \\ 39-21k_1+40k_2&=7t. \hspace{54mm}(\bigstar) \end{align*} Taking $(\bigstar)$ modulo $7$ produces \begin{align*} 4+5k_2&\equiv0\pmod{7} \\ k_2&\equiv2\pmod{7}. \end{align*} Recall that $n>1000.$ From $(2*),$ it follows that \[1063-120k_2<n+63-120k_2=60,\] from which $k_2>8.$ Therefore, the possible values for $k_2$ are $9,16,23,\ldots.$

We need to check whether positive integers $k_1$ and $t$ (where $t\in\{1,2\}$) exist in $(1*):$

  • If $k_2=9,$ then substituting into $(2*)$ gives $n=1077.$ Next, substituting into $(1*)$ produces $1197-63k_1=21t,$ or $57-3k_1=t.$

    There are no solutions $(k_1,t).$

  • If $k_2=16,$ then substituting into $(2*)$ gives $n=1917.$ Next, substituting into $(1*)$ produces $2037-63k_1=21t,$ or $97-3k_1=t.$

    The solution is $(k_1,t)=(32,1).$

Finally, the least such positive integer $n$ is $1917.$ The sum of its digits is $1+9+1+7=\boxed{\textbf{(C) } 18}.$

~MRENTHUSIASM

Solution 10 (Euclidean Algorithm)

Because we are finding value of $n$ for $n > 1000$, let $n = 1000 + k$.

Using the Euclidian Algorithm, \begin{align*} \gcd(63, n+120) &= \gcd(63, 1000 + k + 120) = \gcd(63, k + 1120 - 63 \cdot 18) = \gcd(63, k-14) = 21, \\ \gcd(n+63, 120) &= \gcd(k + 1000 + 63, 120) = \gcd(k + 1000 + 63 - 120 \cdot 9, 120) = \gcd(k-17, 120) = 60. \end{align*} So, we have \begin{align*} k &\equiv 14 \pmod{21}, \\ k &\equiv 17 \pmod{60}. \end{align*} Let $k = 21p + 14 = 60q + 17$, $7p= 20q + 1$, $7p = 21q - (q - 1)$, $q - 1$ is a multiple of $7$.

Let $q-1 = 7r$, $q = 7r+1$, $k = 60(7r+1) + 17 = 420r + 77$.

$n = 1000 + k = 420r + 1077$, $n = 1077, 1497, 1917$.

By substituting $1077$, $1497$, $1917$ into $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60$, $1077$ and $1497$ aren't valid answers, only $1917$ is. Therefore, the answer is $1+9+1+7=\boxed{\textbf{(C) } 18}$.

~isabelchen

Solution 11 (Diophantine Equations)

We know $63 = 3 \cdot 21$, and $\gcd(63, n + 120) = 21 \Longrightarrow n + 120 = 21a$, where $a$ is not a multiple of $3$.

Also, $120 = 2 \cdot 60$, and $gcd(n+63, 120) = 60 \Longrightarrow n + 63 = 60b$, where $b$ is not a multiple of $2$.

Let $n+63 = 60b = x$, $n = x - 63$, $n+120 = x+57 = 21a$.

Now the problem becomes $\gcd(63, x+57) =21$ and $\gcd(x, 120)=60$.

Meaning $x + 57$ has to be a multiple of $21$ but not $63$, and $x$ is a multiple of $60$ but not $120$.

Using trial and error, the least values are $x = 33\cdot60 = 1980$ and $n = x-63 = 1917$.

Therefore, the answer is $1+9+1+7=\boxed{\textbf{(C) } 18}$.

~isabelchen

Solution 12 (Chinese Remainder Theorem)

We have \begin{align*} \gcd(63, n+120) &= 21 \Longrightarrow n + 120 \equiv 0 \pmod{21}, \text{ where } 9 \nmid n + 120, \\ \gcd(n+63, 120) &= 60 \Longrightarrow n + 63 \equiv 0 \pmod{60}, \text{ where } 8 \nmid n + 63. \end{align*} So, we conclude that $n \equiv 6 \pmod{21}$ and $n \equiv 57 \pmod{60}$, respectively.

Because the $2$ moduli $21$ and $60$ are not relatively prime, namely $\gcd{(21, 60)} = 3$, $21 = 3 \cdot 7$, and $60 = 3 \cdot 20$, we convert the system of $2$ linear congruences with non-coprime moduli into a system of $3$ linear congruences with coprime moduli: \begin{align*} n &\equiv 0 \pmod{3}, \\ n &\equiv 6 \pmod{7}, \\ n &\equiv 17 \pmod{20}. \end{align*} By Chinese Remainder Theorem, the general solution of system of $3$ linear congruences is \[n = 420k + 1077.\] We construct the following table: \[\begin{array}{c|c|c|c} n & 1077 & 1497 & 1917\\ \hline n + 120 & 1197 & 1617 & 2037\\ \hline n + 63 & 1140 & 1560 & 1980\\ \end{array}\] Only $n = 1917$ satisfies $9 \nmid n + 120$ and $8 \nmid n + 63$. Therefore, the answer is $1+9+1+7=\boxed{\textbf{(C) } 18}$.

~isabelchen

Video Solutions

Video Solution 1 (Richard Rusczyk)

https://artofproblemsolving.com/videos/amc/2020amc10a/514

Video Solution 2

https://youtu.be/8mNMKH0T9W0 - Happytwin

Video Solution 3 (Quick & Simple)

https://youtu.be/e5BJKMEIPEM

Education The Study of Everything

Video Solution 4

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

Video Solution 5

https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0

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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