Difference between revisions of "2021 Fall AMC 10B Problems/Problem 22"
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==Solution 2 (bash)== | ==Solution 2 (bash)== | ||
+ | Since we have a wonky function, we start by trying out some small cases and see what happens. If <math>j</math> is <math>1</math> and <math>k</math> is <math>2</math>, then there is once case. We have <math>2</math> mod <math>3</math> for this case. If <math>N</math> is <math>3</math>, we have <math>1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3</math> which is still <math>2</math> mod <math>3</math>. If <math>N</math> is <math>4</math>, we have to add <math>1 \cdot 4 + 2 \cdot 4 + 3 \cdot 4</math> which is a multiple of <math>3</math>, meaning that we are still at <math>2</math> mod <math>3</math>. If we try a few more cases, we find that when <math>N</math> is <math>8</math>, we get a multiple of <math>3</math>. When <math>N</math> is <math>9</math>, we are adding <math>0</math> mod <math>3</math>, and therefore, we are still at a multiple of <math>3</math>. | ||
+ | |||
+ | When <math>N</math> is <math>10</math>, then we get <math>0</math> mod <math>3</math> + <math>10(1+2+3+...+9)</math> which is <math>10</math> times a multiple of <math>3</math>. Therefore, we have another multiple of <math>3</math>. When <math>N</math> is <math>11</math>, so we have <math>2</math> mod <math>3</math>. So, every time we have <math>-1</math> mod <math>9</math>, <math>0</math> mod <math>9</math>, and <math>1</math> mod <math>9</math>, we always have a multiple of <math>3</math>. Think about it: When <math>N</math> is <math>1</math>, it will have to be <math>0 \cdot 1</math>, so it is a multiple of <math>3</math>. Therefore, our numbers are <math>8, 9, 10, 17, 18, 19, 26, 27, 28, 35</math>. Adding the numbers up, we get <math>\boxed{(B) 197}</math> | ||
+ | |||
+ | ~Arcticturn | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:33, 23 November 2021
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 once 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
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 |
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