Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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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 + 57</math>, <math>l = 7a + 19</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 = 53</math>, <math>k = 97</math>. So the answer is <math>\boxed{1917}</math>. | 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 + 57</math>, <math>l = 7a + 19</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 = 53</math>, <math>k = 97</math>. So the answer is <math>\boxed{1917}</math>. | ||
==Solution 5== | ==Solution 5== | ||
− | You can first find that n must be congruent to 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</math>=<math>\boxed{1917}</math>. |
== Video Solution == | == Video Solution == |
Revision as of 23:47, 3 February 2020
Contents
[hide]Problem
Let be the least positive integer greater than
for which
What is the sum of the digits of
?
Solution 1
We know that , so we can write
. Simplifying, we get
. Similarly, we can write
, or
. Solving these two modular congruences,
which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than
, we find the least solution is
. However, we are have not considered cases where
or
.
so we try
.
so again we add
to
. It turns out that
does indeed satisfy the original conditions, so our answer is
.
Solution 2 (bashing)
We are given that and
. This tells us that
is divisible by
but not
. It also tells us that
is divisible by 60 but not 120. Starting, we find the least value of
which is divisible by
which satisfies the conditions for
, which is
, making
. We then now keep on adding
until we get a number which satisfies the second equation. This number turns out to be
, whose digits add up to
.
-Midnight
Solution 3 (bashing but worse)
Assume that has 4 digits. Then
, where
,
,
,
represent digits of the number (not to get confused with
). As given the problem,
and
. So we know that
(last digit of
). That means that
and
. We can bash this after this. We just want to find all pairs of numbers
such that
is a multiple of 7 that is
greater than a multiple of
. Our equation for
would be
and our equation for
would be
, where
is any integer. We plug this value in until we get a value of
that makes
satisfy the original problem statement (remember,
). After bashing for hopefully a couple minutes, we find that
works. So
which means that the sum of its digits is
.
~ Baolan
Solution 4
The conditions of the problem reduce to the following. where
and
where
. From these equations, we see that
. Solving this diophantine equation gives us that
,
form. Since,
is greater than
, we can do some bounding and get that
and
. Now we start the bash by plugging in numbers that satisfy these conditions. We get
,
. So the answer is
.
Solution 5
You can first find that n must be congruent toand
. The we can find that
and
, 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
=
.
Video Solution
https://youtu.be/tk3yOGG2K-s -
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 |
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