Difference between revisions of "2021 Fall AMC 10B Problems/Problem 22"
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<cmath>\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.</cmath> | <cmath>\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.</cmath> | ||
− | Clearly, <math> S_2\equiv 2\pmod{3}. </math> Using the above result, we have <math> S_5\equiv 1\pmod{3} </math>, and <math> S_8 </math>, <math> S_9 </math>, and <math> S_{10} </math> are all divisible by <math> 3 </math>. After <math> 3\cdot 3=9 </math>, we have <math> S_{17} </math>, <math> S_{18} </math>, and <math> S_{19} </math> all divisible by <math> 3 </math>, as well as <math> S_{26}, S_{27}, S_{28} </math>, and <math> S_{35} </math>. Thus, our answer is <math> 8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{\mathrm{(B)}\ 197} </math> | + | Clearly, <math> S_2\equiv 2\pmod{3}. </math> Using the above result, we have <math> S_5\equiv 1\pmod{3} </math>, and <math> S_8 </math>, <math> S_9 </math>, and <math> S_{10} </math> are all divisible by <math> 3 </math>. After <math> 3\cdot 3=9 </math>, we have <math> S_{17} </math>, <math> S_{18} </math>, and <math> S_{19} </math> all divisible by <math> 3 </math>, as well as <math> S_{26}, S_{27}, S_{28} </math>, and <math> S_{35} </math>. Thus, our answer is <math> 8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{\mathrm{(B)}\ 197} .</math> -BorealBear |
+ | |||
+ | -minor edit by [[User: Yiyj1|Yiyj1]] | ||
==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 one 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{\textbf{(B) 197}}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Solution 3 == | ||
+ | Denote <math>A_{n, <} = \left\{ \left( j , k \right) : 1 \leq j < k \leq n \right\}</math>, <math>A_{n, >} = \left\{ \left( j , k \right) : 1 \leq k < j \leq n \right\}</math> and <math>A_{n, =} = \left\{ \left( j , k \right) : 1 \leq j = k \leq n \right\}</math>. | ||
+ | |||
+ | Hence, <math>\sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk = S_n</math>. | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | S_n & = \frac{1}{2} \left( \sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk \right) \\ | ||
+ | & = \frac{1}{2} \left( \sum_{1 \leq j, k \leq n} jk - {\left( j , k \right) \in A_{n,=}} jk \right) \\ | ||
+ | & = \frac{1}{2} \left( \sum_{j=1}^n \sum_{k=1}^n jk - \sum_{j=1}^n j^2 \right) \\ | ||
+ | & = \frac{1}{2} \left( \frac{n^2 \left( n + 1 \right)^2}{4} | ||
+ | - \frac{n \left( n + 1 \right) \left( 2 n + 1 \right) }{6} \right) \\ | ||
+ | & = \frac{\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)}{24} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Hence, <math>S_n</math> is divisible by 3 if and only if <math>\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)</math> is divisible by <math>24 \cdot 3 = 8 \cdot 9</math>. | ||
+ | |||
+ | First, <math>\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)</math> is always divisible by 8. Otherwise, <math>S_n</math> is not even an integer. | ||
+ | |||
+ | Second, we find conditions for <math>n</math>, such that <math>\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)</math> is divisible by 9. | ||
+ | |||
+ | Because <math>3 n + 2</math> is not divisible by 3, it cannot be divisible by 9. | ||
+ | |||
+ | Hence, we need to find conditions for <math>n</math>, such that <math>\left( n - 1 \right) n \left( n + 1 \right)</math> is divisible by 9. | ||
+ | This holds of <math>n \equiv 0, \pm 1 \pmod{9}</math>. | ||
+ | |||
+ | Therefore, the 10 least values of <math>n</math> such that <math>\left( n - 1 \right) n \left( n + 1 \right)</math> is divisible by 9 (equivalently, <math>S_n</math> is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35. | ||
+ | Their sum is 197. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) }197}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/_PDIrta6r8s | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution 2 by WhyMath== | ||
+ | https://youtu.be/M8tF1uSOfXA | ||
+ | |||
+ | ~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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:36, 13 November 2023
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
-minor edit by Yiyj1
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.