2020 AMC 10A Problems/Problem 24
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Bashing)
- 5 Solution 4 (Bashing but Worse)
- 6 Solution 5
- 7 Solution 6
- 8 Solution 7 (Reverse Euclidean Algorithm)
- 9 Solution 8
- 10 Solution 9
- 11 Solution 10
- 12 Solution 11 (Euclidean Algorithm)
- 13 Solution 12 (Euclidean Algorithm)
- 14 Solution 13 (Diophantine Equations)
- 15 Video Solutions
- 16 See Also
Problem
Let be the least positive integer greater than
for which
What is the sum of the digits of ?
Solution 1
We know that by the Euclidean Algorithm. Hence, let
. Subtracting the
equations,
. Letting
,
. Taking
, we have
. We are given
. Notice that if
then the condition
is violated. The next possible value of
satisfies the given condition, giving us the answer
. Alternatively, we could have said
for
only, so
, giving us our answer. Since the problem asks for the sum of the digits of
,
=
or
is our answer.
~Prabh1512, with edits by Terribleteeth.
Solution 2
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 the Chinese Remainder Theorem. Now, since the problem is asking for the least positive integer greater than
, we find the least solution is
.
However, we 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 3 (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 4 (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 5
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.
is the first number that works so we get
,
.
. Our answer is then
.
Solution 6
You can first find that n must be congruent to and
. 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
.-happykeeper
Solution 7 (Reverse Euclidean Algorithm)
We are given that and
By applying the Euclidean algorithm, but in reverse, we have
and
We now know that must be divisible by
and
so it is divisible by
Therefore,
for some integer
We know that
or else the first condition won't hold (
will be
) and
or else the second condition won't hold (
will be
). Since
gives us too small of an answer, then
so the answer is
Solution 8
tells us
. The smallest
that satisfies the previous condition and
is
, so we start from there. If
, then
. Because
,
or
. We see that
, which does not fulfill the requirement for
, so we continue by keep on adding
to
, in order to also fulfill the requirement for
. Soon, we see that
decreases by
every time we add
, so we can quickly see that
because at that point
. Adding up all the digits in
, we have
.
-SmileKat32
Solution 9
We are able to set-up the following system-of-congruences:
Therefore, by definition, we are able to set-up the following system of equations:
Thus,
We know
and since
therefore
Simplifying this congruence further, we have
Thus, by definition,
Substituting this back into our original equation,
By definition, we are able to set-up the following congruence:
Thus,
, so our answer is simply
.
(Remarks. since
by definition &
since
by definition.
Remember,
Lastly, the reason why is
would be divisible by
, which is not possible due to the certain condition.)
~ nikenissan
Solution 10
First, we find . We know that it is greater than
, so we first input
. From the first equation,
, we know that if
is correct, after we add
to it, it should be divisible by
, but not
.
Uh oh. To get to the nearest number divisible by
, we have to add
to cancel out the remainder. (Note that we don't subtract
to get to
;
is already at its lowest possible value!)
Adding
to
gives us
. (Note:
is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
Now using, the second equation, , we know that if
is correct, after we add
to it, it should be divisible by
, but not
.
Uh oh (again). This requires some guessing and checking. We can add
over and over again until
is valid. This changes
while also maintaining that
has no remainders.
After adding
once, we get
. By pure luck, adding
two more times gives us
with no remainders.
We now have
. However, this number is divisible by
. To get the next possible number, we add the LCM of
and
(once again, to maintain divisibility), which is
. Unfortunately,
is still divisible by
. Adding
again gives us
, which is valid. However, remember that this is equal to
, so subtracting
from
gives us
, which is
.
The sum of its digits are .
So, our answer is . ~ primegn
Solution 11 (Euclidean Algorithm)
By the Euclidean Algorithm, we have
Clearly,
must be either
or
and
must be
More generally, let so we get
Subtracting
from
and then simplifying give
Taking
modulo
produces
Recall that
From
it follows that
from which
Therefore, the possible values for
are
We need to check whether positive integers and
(where
) exist in
- If
then substituting into
gives
Next, substituting into
produces
or
There are no solutions
- If
then substituting into
gives
Next, substituting into
produces
or
The solution is
Finally, the least such positive integer is
The sum of its digits is
~MRENTHUSIASM
Solution 12 (Euclidean Algorithm)
Because we are finding value of for
, let
.
Using the Euclidian Algorithm,
So, we have
Let
,
,
,
is a multiple of
.
Let ,
,
.
,
.
By substituting ,
,
into
and
,
and
aren't valid answers, only
is. Therefore, the answer is
.
Solution 13 (Diophantine Equations)
We know , and
, where
is not a multiple of
.
Also, , and
, where
is not a multiple of
.
Let ,
,
.
Now the problem becomes and
.
Meaning has to be a multiple of
but not
, and
is a multiple of
but not
.
Using trial and error, the least values are and
.
Therefore, the answer is .
Video Solutions
Video Solution 1 (Richard Rusczyk)
https://artofproblemsolving.com/videos/amc/2020amc10a/514
Video Solution 2
https://youtu.be/8mNMKH0T9W0 - Happytwin
Video Solution 3 (Quick & Simple)
Education The Study of Everything
Video Solution 4
https://www.youtube.com/watch?v=gdGmSyzR908&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=5 ~ MathEx
Video Solution 5
https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0
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 | ||
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