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== | + | == 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. | ||
− | |||
− | + | ~Prabh1512 | |
− | We are given that <math>\gcd(63, n+120)=21< | + | |
+ | |||
+ | |||
+ | |||
+ | |||
+ | == 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>. | ||
+ | |||
+ | ==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>. | ||
-Midnight | -Midnight | ||
− | ==Solution | + | ==Solution 4 (bashing but worse)== |
− | Assume that <math>n< | + | 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 | + | ==Solution 5== |
− | The conditions of the problem reduce to the following. <math>n+120 = 21k< | + | 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}< | + | \boxed{\textbf{(C) } 18}<math>. |
Edited by ~fastnfurious1 | Edited by ~fastnfurious1 | ||
− | ==Solution | + | ==Solution 6== |
− | You can first find that n must be congruent to <math>6\equiv0\pmod {21}< | + | 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 | + | ==Solution 7 (Reverse Euclidean Algorithm)== |
− | We are given that <math>\gcd(63, n+120) =21< | + | 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< | + | 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 | + | ==Solution 8== |
− | <math>\gcd(n+63,120)=60< | + | </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 | + | ==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},< | + | 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.< | + | 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< | + | Thus, </math>n = 1917<math>, so our answer is simply </math>\boxed{18}<math>. |
− | (Remarks. <math>n \equiv 6 \pmod{21}< | + | (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}.< | + | 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< | + | 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 | + | == Solution 10== |
− | First, we find <math>n< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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< | + | The sum of its digits are </math>1 + 9 + 1 + 7 = 18<math>. |
− | So, our answer is <math>\boxed{\textbf{(C) }18} | + | So, our answer is </math>\boxed{\textbf{(C) }18}$. ~ primegn |
== Video Solution 1 == | == Video Solution 1 == |
Revision as of 08:05, 25 December 2020
Problem
Let be the least positive integer greater than for whichWhat is the sum of the digits of ?
Solution 1
We know that . Hence, . Subtracting the equations, . Letting , . Taking \alpha \equiv{14} \pmod{40}n=21\alpha -57 >1000 \implies \alpha \geq 51\alpha =54(\alpha,3)=1\alpha = 94\boxed{1917}\alpha = 40k+14 \equiv{0} \pmod{3}k \equiv{1} \pmod{3}k \equiv{0,2} \pmod{3}$, giving us our answer.
~Prabh1512
== Solution 2==
We know that$ (Error compiling LaTeX. ! Missing $ inserted.)gcd(63, n+120)=21n+120\equiv0\pmod {21}n\equiv6\pmod {21}n+63\equiv0\pmod {60}n\equiv-3\pmod {60}n\equiv237\pmod {420}1000n=1077gcd(63, n+120) =63gcd(n+63, 120) =120{1077+120}\equiv0\pmod {63}n=1077+420=1497{1497+63}\equiv0\pmod {120}420nn=1497+420=19171+9+1+7=\boxed{\textbf{(C) }18}$.
==Solution 3 (bashing)==
We are given that$ (Error compiling LaTeX. ! Missing $ inserted.)\gcd(63, n+120)=21\gcd(n+63,120) = 60n+1202163n+63n+12021n1134n=1014211917\boxed{\textbf{(C) } 18}$.
-Midnight
==Solution 4 (bashing but worse)==
Assume that$ (Error compiling LaTeX. ! Missing $ inserted.)nn = abcdabcda * b * c * dgcd(63, n + 120) = 21gcd(n + 63, 120) = 60d = 7n12 + abc \equiv0\pmod {7}7 + abc\equiv0\pmod {6}(x, y)x5612 + abc42*j + 35 = x7 + abc 42*j + 30 = yjabcn = abc7abc > 100abc = 191n = 1917\boxed{\textbf{(C) } 18}$.
~ Baolan
==Solution 5== The conditions of the problem reduce to the following.$ (Error compiling LaTeX. ! Missing $ inserted.)n+120 = 21kgcd(k,3) = 1n+63 = 60lgcd(l,2) = 121k - 60l = 57k = 20a + 17l = 7a + 5n1000k > 53l > 17l = 33k = 971917 \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}57\equiv0\pmod {60}n=21x+6n=60y+571+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\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+1832160,\text{lcm}(21, 60) = 420.n+183 = 420kk.3 \nmid k,\gcd632 \nmid k,\gcd120k = 1k=5 \implies n = 1917,1+9+1+7 = \boxed{\textbf{(C) } 18}.\gcd(n+63,120)=60n+63\equiv60\pmod {120}n+63n>10001140n+63=1140n+120=1197\gcd(n+120,63)=21n+120\equiv21\pmod {63}n+120\equiv42\pmod {63}1197\equiv0\pmod {63}n+1201201197n+63n+120\pmod {63}6120n=1917n+120\equiv21\pmod {63}1917\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},7a = 20b + 17,20b + 17 \equiv 0 \pmod{7}.b = 7x + 3.n = 1917\boxed{18}$.
(Remarks.$ (Error compiling LaTeX. ! Missing $ inserted.)n \equiv 6 \pmod{21}n \equiv -120 \pmod{21},$by definition &$ (Error compiling LaTeX. ! Misplaced alignment tab character &.)n \equiv 57 \pmod{60}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}.n \neq 1077n + 12063$, which is not possible due to the certain condition.)
~ nikenissan
== Solution 10==
First, we find$ (Error compiling LaTeX. ! Missing $ inserted.)n1000n = 1000gcd(63, n + 120) = 21n12021632114753n141000n = 1014n$is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
Now using, the second equation,$ (Error compiling LaTeX. ! Missing $ inserted.)gcd(n + 63, 120) = 60n636012021nn\frac{n + 120}{21}2118 r 1821191077 + 21 + 21 + 21 = 114012021604201140 + 420 = 15601204201980n + 636319801917n$.
The sum of its digits are$ (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
Video Solution 3
https://www.youtube.com/watch?v=gdGmSyzR908&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=5 ~ MathEx
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.