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

Revision as of 01:28, 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 $(n+57,63)=21, (n-57, 120)= 60$ by the Euclidean Algorithm. Hence, let $n+57=21\alpha, n-57=60 \gamma, (\alpha,3)=1, (\gamma,2)=1$ . Subtracting the $2$ equations, $38=7\alpha-20\gamma$ . Letting $\gamma = 2s+1$ , $58=7\alpha-40s$ . Taking $\mod{40}$ , we have $\alpha \equiv{14} \pmod{40}$ . We are given $n=21\alpha -57 >1000 \implies \alpha \geq 51$ . Notice that if $\alpha =54$ then the condition $(\alpha,3)=1$ is violated. The next possible value of $\alpha = 94$ satisfies the given condition, giving us the answer $\boxed{1917}$ . 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 $n$ , $1+9+1+7$ = $18$ or $\boxed{\textbf{(C) } 18}$ is our answer.

~Prabh1512, with edits by Terribleteeth.

Solution 2

We know that $\text{gcd}(63, n+120)=21$ , 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 the 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 have not considered cases where $\text{gcd}(63, n+120)=63$ or $\text{gcd}(n+63, 120)=120$ .

${1077+120} \equiv 0 \pmod {63}$ 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 3 (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 4 (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 5

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. $a=4$ is the first number that works so we get $l = 33$ , $k = 97$ . $n=21(97)-120=60(33)-63=1917$ . Our answer is then $1+9+1+7=\boxed{\textbf{(C) } 18}$ .

Solution 6

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 7 (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 8

$\gcd(n+63,120)=60$ 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 $\boxed{\textbf{(C) } 18}$ .

-SmileKat32

Solution 9

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, $n = 1917$ , 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

Solution 10

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 $63$ . $\frac{n + 120}{21},$ $\frac{1120}{21},$ $53 r 7.$ Uh oh. 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, after we add $63$ to it, it should be divisible by $60$ , but not $120$ . $\frac{n + 63}{60},$ $\frac{1077}{60},$ $17r57.$ Uh oh (again). 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 $1 + 9 + 1 + 7 = 18$ .

So, our answer is $\boxed{\textbf{(C) }18}$ . ~ primegn

Solution 11 (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 12 (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 13 (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 14 (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

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

@@ Line 178: / Line 178: @@
 == Solution 14 (Chinese Remainder Theorem) ==
+We have
+<cmath>\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*}</cmath>
+So, we conclude that <math>n \equiv 6 \pmod{21}</math> and <math>n \equiv 57 \pmod{60}</math>, respectively.
-<math>\gcd(63, n+120) =21 \Longrightarrow n + 120 \equiv 0 (\bmod \quad 21), \quad 9 \nmid n + 120</math>
+Because the <math>2</math> moduli <math>21</math> and <math>60</math> are not relatively prime, namely <math>\gcd{(21, 60)} = 3</math>, <math>21 = 3 \cdot 7</math>, and <math>60 = 3 \cdot 20</math>, we convert the system of <math>2</math> linear congruences with non-coprime moduli into a system of <math>3</math> linear congruences with coprime moduli:
+<cmath>\begin{align*}
-<math>\gcd(n+63, 120) = 60 \Longrightarrow n + 63 \equiv 0 (\bmod \quad 60), \quad 8 \nmid n + 63</math>
+n &\equiv 0 \pmod{3}, \\
+n &\equiv 6 \pmod{7}, \\
-<math>n \equiv 6 (\bmod \quad 21)</math>, <math>n \equiv 57 (\bmod \quad 60)</math>
+n &\equiv 17 \pmod{20}.
+\end{align*}</cmath>
-Because the <math>2</math> moduli <math>21</math> and <math>60</math> are not relatively prime, <math>\gcd{(21, 60)} = 3</math>, <math>21 = 3 \cdot 7</math>, <math>60 = 3 \cdot 20</math>, we convert the system of <math>2</math> linear congruences with non-coprime moduli into a system of <math>3</math> linear congruences with coprime moduli:
+By [https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem Chinese Remainder Theorem], the general solution of system of <math>3</math> linear congruences is <cmath>n = 420k + 1077.</cmath>
+We construct the following table:
-<math>n \equiv 0 (\bmod \quad 3)</math>, <math>n \equiv 6 (\bmod \quad 7)</math>, <math>n \equiv 17 (\bmod \quad 20)</math>
-By [https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem Chinese Remainder Theorem], the general solution of system of <math>3</math> linear congruences is:
- <math>n \equiv r_1 (\bmod { \quad m_1})</math>, <math>n \equiv r_2 (\bmod { \quad m_2})</math>, <math>n \equiv r_3 (\bmod { \quad m_3})</math>, <math>(m_1, m_2) = 1</math>, <math>(m_2, m_3) = 1</math>, <math>(m_3, m_1) = 1</math>
- Find <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> such that <math>k_1 m_2 m_3 \equiv 1 (\bmod{ \quad m_1})</math>, <math>k_2 m_3 m_1 \equiv 1 (\bmod{ \quad m_2})</math>, <math>k_3 m_1 m_2 \equiv 1 (\bmod{ \quad m_3})</math>
- Then <math>n \equiv k_1 m_2 m_3 r_1 + k_2 m_3 m_1 r_2 + k_3 m_1 m_2 r_3 (\bmod{ \quad m_1 m_2 m_3})</math>
-In this problem:
- <math>n \equiv 0 (\bmod { \quad 3})</math>, <math>n \equiv 6 (\bmod \quad 7)</math>, <math>n \equiv 17 (\bmod \quad 20)</math>
- <math>k_1 \cdot 7 \cdot 20\equiv 1 (\bmod{ \quad 3})</math>, <math>k_2 \cdot 3 \cdot 20 \equiv 1 (\bmod{ \quad 7})</math>, <math>k_3 \cdot 3 \cdot 7 \equiv 1 (\bmod{ \quad 20})</math>
- <math>k_1 \equiv 2 (\bmod{ \quad 3})</math>, <math>k_2 \equiv 2 (\bmod{ \quad 7})</math>, <math>k_3 \equiv 1 (\bmod{ \quad 20})</math>,
- Then <math>n \equiv 2 \cdot 3 \cdot 20 \cdot 6 + 1 \cdot 3 \cdot 7 \cdot 17 \equiv 1077 (\bmod{ \quad 420})</math>
-<math>n = 420k + 1077</math>
 <cmath>\begin{array}{c|c|c|c}
 n & 1077 & 1497 & 1917\\
@@ Line 209: / Line 200: @@
 n + 63 & 1140 & 1560 & 1980\\
 \end{array}</cmath>
 Only <math>n = 1917</math> satisfies <math>9 \nmid n + 120</math> and <math>8 \nmid n + 63</math>. Therefore, the answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>.

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

Page

Toolbox

Search

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

Revision as of 01:28, 14 January 2022

Contents

Problem

Solution 1

Solution 2

Solution 3 (Bashing)

Solution 4 (Bashing but Worse)

Solution 5

Solution 6

Solution 7 (Reverse Euclidean Algorithm)

Solution 8

Solution 9

Solution 10

Solution 11 (Euclidean Algorithm)

Solution 12 (Euclidean Algorithm)

Solution 13 (Diophantine Equations)

Solution 14 (Chinese Remainder Theorem)

Video Solutions

Video Solution 1 (Richard Rusczyk)

Video Solution 2

Video Solution 3 (Quick & Simple)

Video Solution 4

Video Solution 5

See Also