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

Revision as of 07:06, 25 December 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 $(n+57,63)=21, (n-57, 120)= 60$ . Hence, $n+57=21\alpha,n-57=60 \gamma, (\alpha,3)=1, (\gamma,2)=1$ . 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.

~Prabh1512

Solution 2

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 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. We get $l = 33$ , $k = 97$ . So the answer is $1917 \implies 1+9+1+7= \boxed{\textbf{(C) } 18}$ .

Edited by ~fastnfurious1

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

Video Solution 1

https://youtu.be/tk3yOGG2K-s ~ Richard Rusczyk

Video Solution 2

Education The Study of Everything

https://youtu.be/e5BJKMEIPEM

Video Solution 3

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

@@ Line 6: / Line 6: @@
 == Solution 1 ==
-We know that <math>(n+57,63)=21, (n-57, 120)= 60</math>. Hence, <math>n+57=21\alpha,n-57=60 \gamma, (\alpha,3)=1, (\gamma,2)=1</math>. Subtracting the <math>2</math> equations, <math>38=7\alpha-20\gamma</math>. Letting <math>\gamma = 2s+1</math>, <math>58=7\alpha-40s</math>. Taking <math>\mod{40}, we have </math>\alpha \equiv{14} \pmod{40}<math>. We are given </math>n=21\alpha -57 >1000 \implies \alpha \geq 51<math>. Notice that if </math>\alpha =54<math> then the condition </math>(\alpha,3)=1<math> is violated. The next possible value of  </math>\alpha = 94<math> satisfies the given condition, giving us the answer </math>\boxed{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.
+We know that <math>(n+57,63)=21, (n-57, 120)= 60</math>. Hence, <math>n+57=21\alpha,n-57=60 \gamma, (\alpha,3)=1, (\gamma,2)=1</math>. Subtracting the <math>2</math> equations, <math>38=7\alpha-20\gamma</math>. Letting <math>\gamma = 2s+1</math>, <math>58=7\alpha-40s</math>. Taking <math>\mod{40}</math>, we have <math>\alpha \equiv{14} \pmod{40}</math>. We are given <math>n=21\alpha -57 >1000 \implies \alpha \geq 51</math>. Notice that if <math>\alpha =54</math> then the condition <math>(\alpha,3)=1</math> is violated. The next possible value of  <math>\alpha = 94</math> satisfies the given condition, giving us the answer <math>\boxed{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.
@@ Line 18: / Line 18: @@
 == Solution 2==
-We know that </math>gcd(63, n+120)=21<math>, so we can write </math>n+120\equiv0\pmod {21}<math>. Simplifying, we get </math>n\equiv6\pmod {21}<math>. Similarly, we can write </math>n+63\equiv0\pmod {60}<math>, or </math>n\equiv-3\pmod {60}<math>. Solving these two modular congruences, </math>n\equiv237\pmod {420}<math> 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 </math>1000<math>, we find the least solution is </math>n=1077<math>. However, we are have not considered cases where </math>gcd(63, n+120) =63<math> or </math>gcd(n+63, 120) =120<math>. </math>{1077+120}\equiv0\pmod {63}<math> so we try </math>n=1077+420=1497<math>. </math>{1497+63}\equiv0\pmod {120}<math> so again we add </math>420<math> to </math>n<math>. It turns out that </math>n=1497+420=1917<math> does indeed satisfy the original conditions, so our answer is </math>1+9+1+7=\boxed{\textbf{(C) }18}<math>.
+We know that <math>gcd(63, n+120)=21</math>, so we can write <math>n+120\equiv0\pmod {21}</math>. Simplifying, we get <math>n\equiv6\pmod {21}</math>. Similarly, we can write <math>n+63\equiv0\pmod {60}</math>, or <math>n\equiv-3\pmod {60}</math>. Solving these two modular congruences, <math>n\equiv237\pmod {420}</math> 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 <math>1000</math>, we find the least solution is <math>n=1077</math>. However, we are have not considered cases where <math>gcd(63, n+120) =63</math> or <math>gcd(n+63, 120) =120</math>. <math>{1077+120}\equiv0\pmod {63}</math> so we try <math>n=1077+420=1497</math>. <math>{1497+63}\equiv0\pmod {120}</math> so again we add <math>420</math> to <math>n</math>. It turns out that <math>n=1497+420=1917</math> does indeed satisfy the original conditions, so our answer is <math>1+9+1+7=\boxed{\textbf{(C) }18}</math>.
 ==Solution 3 (bashing)==
-We are given that </math>\gcd(63, n+120)=21<math> and </math>\gcd(n+63,120) = 60<math>. This tells us that </math>n+120<math> is divisible by </math>21<math> but not </math>63<math>. It also tells us that </math>n+63<math> is divisible by 60 but not 120. Starting, we find the least value of </math>n+120<math> which is divisible by </math>21<math> which satisfies the conditions for </math>n<math>, which is </math>1134<math>, making </math>n=1014<math>. We then now keep on adding </math>21<math> until we get a number which satisfies the second equation. This number turns out to be </math>1917<math>, whose digits add up to </math>\boxed{\textbf{(C) } 18}<math>.
+We are given that <math>\gcd(63, n+120)=21</math> and <math>\gcd(n+63,120) = 60</math>. This tells us that <math>n+120</math> is divisible by <math>21</math> but not <math>63</math>. It also tells us that <math>n+63</math> is divisible by 60 but not 120. Starting, we find the least value of <math>n+120</math> which is divisible by <math>21</math> which satisfies the conditions for <math>n</math>, which is <math>1134</math>, making <math>n=1014</math>. We then now keep on adding <math>21</math> until we get a number which satisfies the second equation. This number turns out to be <math>1917</math>, whose digits add up to <math>\boxed{\textbf{(C) } 18}</math>.
 -Midnight
@@ Line 28: / Line 28: @@
 ==Solution 4 (bashing but worse)==
-Assume that </math>n<math> has 4 digits. Then </math>n = abcd<math>, where </math>a<math>, </math>b<math>, </math>c<math>, </math>d<math> represent digits of the number (not to get confused with </math>a * b * c * d<math>). As given the problem, </math>gcd(63, n + 120) = 21<math> and </math>gcd(n + 63, 120) = 60<math>. So we know that </math>d = 7<math> (last digit of </math>n<math>). That means that </math>12 + abc \equiv0\pmod {7}<math> and </math>7 + abc\equiv0\pmod {6}<math>. We can bash this after this. We just want to find all pairs of numbers </math>(x, y)<math> such that </math>x<math> is a multiple of 7 that is </math>5<math> greater than a multiple of </math>6<math>. Our equation for </math>12 + abc<math> would be </math>42*j + 35 = x<math> and our equation for </math>7 + abc<math> would be </math> 42*j + 30 = y<math>, where </math>j<math> is any integer. We plug this value in until we get a value of </math>abc<math> that makes </math>n = abc7<math> satisfy the original problem statement (remember, </math>abc > 100<math>). After bashing for hopefully a couple minutes, we find that </math>abc = 191<math> works. So </math>n = 1917<math> which means that the sum of its digits is </math>\boxed{\textbf{(C) } 18}<math>.
+Assume that <math>n</math> has 4 digits. Then <math>n = abcd</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> represent digits of the number (not to get confused with <math>a * b * c * d</math>). As given the problem, <math>gcd(63, n + 120) = 21</math> and <math>gcd(n + 63, 120) = 60</math>. So we know that <math>d = 7</math> (last digit of <math>n</math>). That means that <math>12 + abc \equiv0\pmod {7}</math> and <math>7 + abc\equiv0\pmod {6}</math>. We can bash this after this. We just want to find all pairs of numbers <math>(x, y)</math> such that <math>x</math> is a multiple of 7 that is <math>5</math> greater than a multiple of <math>6</math>. Our equation for <math>12 + abc</math> would be <math>42*j + 35 = x</math> and our equation for <math>7 + abc</math> would be <math> 42*j + 30 = y</math>, where <math>j</math> is any integer. We plug this value in until we get a value of <math>abc</math> that makes <math>n = abc7</math> satisfy the original problem statement (remember, <math>abc > 100</math>). After bashing for hopefully a couple minutes, we find that <math>abc = 191</math> works. So <math>n = 1917</math> which means that the sum of its digits is <math>\boxed{\textbf{(C) } 18}</math>.
 ~ Baolan
 ==Solution 5==
-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=
+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>.
+\boxed{\textbf{(C) } 18}</math>.
 Edited by ~fastnfurious1
 ==Solution 6==
-You can first find that n must be congruent to </math>6\equiv0\pmod {21}<math> and </math>57\equiv0\pmod {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 = \boxed{\textbf{(C) } 18}<math>.-happykeeper
+You can first find that n must be congruent to <math>6\equiv0\pmod {21}</math> and <math>57\equiv0\pmod {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 = \boxed{\textbf{(C) } 18}</math>.-happykeeper
 ==Solution 7 (Reverse Euclidean Algorithm)==
-We are given that </math>\gcd(63, n+120) =21<math> and </math>\gcd(n+63, 120)=60.<math> By applying the Euclidean algorithm, but in reverse, we have <cmath>\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21</cmath> and <cmath>\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.</cmath>
+We are given that <math>\gcd(63, n+120) =21</math> and <math>\gcd(n+63, 120)=60.</math> By applying the Euclidean algorithm, but in reverse, we have <cmath>\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21</cmath> and <cmath>\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.</cmath>
-We now know that </math>n+183<math> must be divisible by </math>21<math> and </math>60,<math> so it is divisible by </math>\text{lcm}(21, 60) = 420.<math> Therefore, </math>n+183 = 420k<math> for some integer </math>k.<math> We know that </math>3 \nmid k,<math> or else the first condition won't hold (</math>\gcd<math> will be </math>63<math>) and </math>2 \nmid k,<math> or else the second condition won't hold (</math>\gcd<math> will be </math>120<math>). Since </math>k = 1<math> gives us too small of an answer, then </math>k=5 \implies n = 1917,<math> so the answer is </math>1+9+1+7 = \boxed{\textbf{(C) } 18}.<math>
+We now know that <math>n+183</math> must be divisible by <math>21</math> and <math>60,</math> so it is divisible by <math>\text{lcm}(21, 60) = 420.</math> Therefore, <math>n+183 = 420k</math> for some integer <math>k.</math> We know that <math>3 \nmid k,</math> or else the first condition won't hold (<math>\gcd</math> will be <math>63</math>) and <math>2 \nmid k,</math> or else the second condition won't hold (<math>\gcd</math> will be <math>120</math>). Since <math>k = 1</math> gives us too small of an answer, then <math>k=5 \implies n = 1917,</math> so the answer is <math>1+9+1+7 = \boxed{\textbf{(C) } 18}.</math>
 ==Solution 8==
-</math>\gcd(n+63,120)=60<math> tells us </math>n+63\equiv60\pmod {120}<math>. The smallest </math>n+63<math> that satisfies the previous condition and </math>n>1000<math> is </math>1140<math>, so we start from there. If </math>n+63=1140<math>, then </math>n+120=1197<math>. Because </math>\gcd(n+120,63)=21<math>, </math>n+120\equiv21\pmod {63}<math> or </math>n+120\equiv42\pmod {63}<math>. We see that </math>1197\equiv0\pmod {63}<math>, which does not fulfill the requirement for </math>n+120<math>, so we continue by keep on adding </math>120<math> to </math>1197<math>, in order to also fulfill the requirement for </math>n+63<math>. Soon, we see that </math>n+120\pmod {63}<math> decreases by </math>6<math> every time we add </math>120<math>, so we can quickly see that </math>n=1917<math> because at that point </math>n+120\equiv21\pmod {63}<math>. Adding up all the digits in </math>1917<math>, we have </math>\boxed{\textbf{(C) } 18}<math>.
+<math>\gcd(n+63,120)=60</math> tells us <math>n+63\equiv60\pmod {120}</math>. The smallest <math>n+63</math> that satisfies the previous condition and <math>n>1000</math> is <math>1140</math>, so we start from there. If <math>n+63=1140</math>, then <math>n+120=1197</math>. Because <math>\gcd(n+120,63)=21</math>, <math>n+120\equiv21\pmod {63}</math> or <math>n+120\equiv42\pmod {63}</math>. We see that <math>1197\equiv0\pmod {63}</math>, which does not fulfill the requirement for <math>n+120</math>, so we continue by keep on adding <math>120</math> to <math>1197</math>, in order to also fulfill the requirement for <math>n+63</math>. Soon, we see that <math>n+120\pmod {63}</math> decreases by <math>6</math> every time we add <math>120</math>, so we can quickly see that <math>n=1917</math> because at that point <math>n+120\equiv21\pmod {63}</math>. Adding up all the digits in <math>1917</math>, we have <math>\boxed{\textbf{(C) } 18}</math>.
 -SmileKat32
@@ Line 61: / Line 61: @@
 <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
+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,
+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>
@@ Line 70: / Line 70: @@
 By definition, we are able to set-up the following congruence:
 <cmath>n \equiv 237 \pmod{420}.</cmath>
-Thus, </math>n = 1917<math>, so our answer is simply </math>\boxed{18}<math>.
+Thus, <math>n = 1917</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.
+(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>
+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.)
+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
@@ Line 82: / Line 82: @@
 == Solution 10==
-First, we find </math>n<math>. We know that it is greater than </math>1000<math>, so we first input </math>n = 1000<math>. From the first equation, </math>gcd(63, n + 120) = 21<math>, we know that if </math>n<math> is correct, after we add </math>120<math> to it, it should be divisible by </math>21<math>, but not </math>63<math>.
+First, we find <math>n</math>. We know that it is greater than <math>1000</math>, so we first input <math>n = 1000</math>. From the first equation, <math>gcd(63, n + 120) = 21</math>, we know that if <math>n</math> is correct, after we add <math>120</math> to it, it should be divisible by <math>21</math>, but not <math>63</math>.
 <cmath>\frac{n + 120}{21}, </cmath>
 <cmath>\frac{1120}{21}, </cmath>
 <cmath>53 r 7. </cmath>
-Uh oh. To get to the nearest number divisible by </math>21<math>, we have to add </math>14<math> to cancel out the remainder. (Note that we don't subtract </math>7<math> to get to </math>53<math>; </math>n<math> is already at its lowest possible value!)
+Uh oh. To get to the nearest number divisible by <math>21</math>, we have to add <math>14</math> to cancel out the remainder. (Note that we don't subtract <math>7</math> to get to <math>53</math>; <math>n</math> is already at its lowest possible value!)
-Adding </math>14<math> to </math>1000<math> gives us </math>n = 1014<math>. (Note: </math>n<math> is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
+Adding <math>14</math> to <math>1000</math> gives us <math>n = 1014</math>. (Note: <math>n</math> is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
-Now using, the second equation, </math>gcd(n + 63, 120) = 60<math>, we know that if </math>n<math> is correct, after we add </math>63<math> to it, it should be divisible by </math>60<math>, but not </math>120<math>.
+Now using, the second equation, <math>gcd(n + 63, 120) = 60</math>, we know that if <math>n</math> is correct, after we add <math>63</math> to it, it should be divisible by <math>60</math>, but not <math>120</math>.
 <cmath>\frac{n + 63}{60}, </cmath>
 <cmath>\frac{1077}{60}, </cmath>
 <cmath>17r57. </cmath>
-Uh oh (again). This requires some guessing and checking. We can add </math>21<math> over and over again until </math>n<math> is valid. This changes </math>n<math> while also maintaining that </math>\frac{n + 120}{21}<math> has no remainders.
+Uh oh (again). This requires some guessing and checking. We can add <math>21</math> over and over again until <math>n</math> is valid. This changes <math>n</math> while also maintaining that <math>\frac{n + 120}{21}</math> has no remainders.
-After adding </math>21<math> once, we get </math>18 r 18<math>. By pure luck, adding </math>21<math> two more times gives us </math>19<math> with no remainders.
+After adding <math>21</math> once, we get <math>18 r 18</math>. By pure luck, adding <math>21</math> two more times gives us <math>19</math> with no remainders.
-We now have </math>1077 + 21 + 21 + 21 = 1140<math>. However, this number is divisible by </math>120<math>. To get the next possible number, we add the LCM of </math>21<math> and </math>60<math> (once again, to maintain divisibility), which is </math>420<math>. Unfortunately, </math>1140 + 420 = 1560<math> is still divisible by </math>120<math>. Adding </math>420<math> again gives us </math>1980<math>, which is valid. However, remember that this is equal to </math>n + 63<math>, so subtracting </math>63<math> from </math>1980<math> gives us </math>1917<math>, which is </math>n<math>.
+We now have <math>1077 + 21 + 21 + 21 = 1140</math>. However, this number is divisible by <math>120</math>. To get the next possible number, we add the LCM of <math>21</math> and <math>60</math> (once again, to maintain divisibility), which is <math>420</math>. Unfortunately, <math>1140 + 420 = 1560</math> is still divisible by <math>120</math>. Adding <math>420</math> again gives us <math>1980</math>, which is valid. However, remember that this is equal to <math>n + 63</math>, so subtracting <math>63</math> from <math>1980</math> gives us <math>1917</math>, which is <math>n</math>.
-The sum of its digits are </math>1 + 9 + 1 + 7 = 18<math>.
+The sum of its digits are <math>1 + 9 + 1 + 7 = 18</math>.
-So, our answer is </math>\boxed{\textbf{(C) }18}$. ~ primegn
+So, our answer is <math>\boxed{\textbf{(C) }18}</math>. ~ primegn
 == Video Solution 1 ==

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