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

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<math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math>
 
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math>
  
== Solution 1==
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== Solution 1 ==  
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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>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 2 (bashing)==
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~Prabh1512
  
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>.
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== Solution 2==
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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>.
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==Solution 3 (bashing)==
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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
 
-Midnight
  
==Solution 3 (bashing but worse)==
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==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>.
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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
 
~ Baolan
  
==Solution 4==
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==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=
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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 + 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>.
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\boxed{\textbf{(C) } 18}<math>.
  
 
Edited by ~fastnfurious1
 
Edited by ~fastnfurious1
  
==Solution 5==
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==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
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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 6 (Reverse Euclidean Algorithm)==
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==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>
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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>
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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 7==
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==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>.
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</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
 
-SmileKat32
  
==Solution 8==
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==Solution 9==
 
We are able to set-up the following system-of-congruences:  
 
We are able to set-up the following system-of-congruences:  
 
<cmath>n \equiv 6 \pmod {21},</cmath>  
 
<cmath>n \equiv 6 \pmod {21},</cmath>  
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<cmath>21a + 6 = 60b + 57</cmath>
 
<cmath>21a + 6 = 60b + 57</cmath>
 
<cmath>\implies 7a + 2 = 20b + 19.</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  
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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>5b \equiv 1 \pmod{7}</cmath>
 
<cmath>\implies b \equiv 3 \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,  
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Thus, by definition, </math>b = 7x + 3.<math> Substituting this back into our original equation,  
 
<cmath>n = 60(7x + 3) + 57</cmath>
 
<cmath>n = 60(7x + 3) + 57</cmath>
 
<cmath>\implies n = 420x + 180 + 57</cmath>
 
<cmath>\implies n = 420x + 180 + 57</cmath>
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By definition, we are able to set-up the following congruence:
 
By definition, we are able to set-up the following congruence:
 
<cmath>n \equiv 237 \pmod{420}.</cmath>
 
<cmath>n \equiv 237 \pmod{420}.</cmath>
Thus, <math>n = 1917</math>, so our answer is simply <math>\boxed{18}</math>.
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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.  
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(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>
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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.)  
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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
 
~ nikenissan
  
== Solution 9==
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== 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>.  
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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{n + 120}{21}, </cmath>
 
<cmath>\frac{1120}{21}, </cmath>
 
<cmath>\frac{1120}{21}, </cmath>
 
<cmath>53 r 7. </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!)
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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.)
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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>.
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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{n + 63}{60}, </cmath>
 
<cmath>\frac{1077}{60}, </cmath>
 
<cmath>\frac{1077}{60}, </cmath>
 
<cmath>17r57. </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.  
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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.  
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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>.  
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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>.
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The sum of its digits are </math>1 + 9 + 1 + 7 = 18<math>.
  
So, our answer is <math>\boxed{\textbf{(C) }18}</math>. ~ primegn  
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So, our answer is </math>\boxed{\textbf{(C) }18}$. ~ primegn  
  
 
== Video Solution 1 ==
 
== Video Solution 1 ==

Revision as of 08:05, 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$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)\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$ (Error compiling LaTeX. ! Missing $ inserted.)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.$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)\gcd(63, n+120) =21$and$\gcd(n+63, 120)=60.$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$ (Error compiling LaTeX. ! Missing $ inserted.)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: <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$ (Error compiling LaTeX. ! Missing $ inserted.)7a \equiv 0 \pmod {7},$and since$7a = 20b + 17,$therefore$20b + 17 \equiv 0 \pmod{7}.$Simplifying this congruence further, we have  <cmath>5b \equiv 1 \pmod{7}</cmath> <cmath>\implies b \equiv 3 \pmod {7}.</cmath> Thus, by definition,$b = 7x + 3.$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,$n = 1917$, so our answer is simply$\boxed{18}$.

(Remarks.$ (Error compiling LaTeX. ! Missing $ inserted.)n \equiv 6 \pmod{21}$since$n \equiv -120 \pmod{21},$by definition &$ (Error compiling LaTeX. ! Misplaced alignment tab character &.)n \equiv 57 \pmod{60}$since$n \equiv -63 \pmod{60},$by definition.

Remember,$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)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$63$.  <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$21$, 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,$ (Error compiling LaTeX. ! Missing $ inserted.)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$120$. <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$21$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$ (Error compiling LaTeX. ! Missing $ inserted.)1 + 9 + 1 + 7 = 18$.

So, our answer is$ (Error compiling LaTeX. ! Missing $ inserted.)\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

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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