Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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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>a + | + | 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>). So <math>abc > 100</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. After bashing for hopefully just 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>\box{text\{(C) }\ 18}</math>. |
+ | |||
+ | ~ Baolan | ||
== Video Solution == | == Video Solution == |
Revision as of 01:53, 2 February 2020
Contents
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
). So
. 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. After bashing for hopefully just a couple minutes, we find that
works. So
which means that the sum of its digits is $\box{text\{(C) }\ 18}$ (Error compiling LaTeX. Unknown error_msg).
~ Baolan
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.