Difference between revisions of "2021 Fall AMC 10B Problems/Problem 22"
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~Interstigation | ~Interstigation | ||
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+ | ==Video Solution 2 by WhyMath== | ||
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+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2021 Fall|ab=B|num-b=21|num-a=23}} | ||
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Revision as of 08:23, 26 November 2022
Contents
Problem
For each integer , let
be the sum of all products
, where
and
are integers and
. What is the sum of the 10 least values of
such that
is divisible by
?
Solution 1
To get from to
, we add
.
Now, we can look at the different values of mod
. For
and
, then we have
. However, for
, we have
Clearly, Using the above result, we have
, and
,
, and
are all divisible by
. After
, we have
,
, and
all divisible by
, as well as
, and
. Thus, our answer is
. -BorealBear
Solution 2 (bash)
Since we have a wonky function, we start by trying out some small cases and see what happens. If is
and
is
, then there is one case. We have
mod
for this case. If
is
, we have
which is still
mod
. If
is
, we have to add
which is a multiple of
, meaning that we are still at
mod
. If we try a few more cases, we find that when
is
, we get a multiple of
. When
is
, we are adding
mod
, and therefore, we are still at a multiple of
.
When is
, then we get
mod
+
which is
times a multiple of
. Therefore, we have another multiple of
. When
is
, so we have
mod
. So, every time we have
mod
,
mod
, and
mod
, we always have a multiple of
. Think about it: When
is
, it will have to be
, so it is a multiple of
. Therefore, our numbers are
. Adding the numbers up, we get
~Arcticturn
Solution 3
Denote ,
and
.
Hence, .
Therefore,
Hence, is divisible by 3 if and only if
is divisible by
.
First, is always divisible by 8. Otherwise,
is not even an integer.
Second, we find conditions for , such that
is divisible by 9.
Because is not divisible by 3, it cannot be divisible by 9.
Hence, we need to find conditions for , such that
is divisible by 9.
This holds of
.
Therefore, the 10 least values of such that
is divisible by 9 (equivalently,
is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35.
Their sum is 197.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
~Interstigation
Video Solution 2 by WhyMath
~savannahsolver
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.