2020 AMC 10A Problems/Problem 24

Revision as of 08:05, 25 December 2020 by Prabh1512 (talk | contribs)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS