Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 10B Problems/Problem 22"

(→‎Problem)
(→‎Solution 2 (bash))
Line 13: Line 13:
  
 
==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 10:33, 23 November 2021

Problem

For each integer $n\geq 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1\leq j<k\leq n$. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$?

$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$

Solution 1

To get from $S_n$ to $S_{n+1}$, we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1)^2}{2}$.

Now, we can look at the different values of $n$ mod $3$. For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$, then we have $\frac{n(n+1)^2}{2}\equiv 0\pmod{3}$. However, for $n\equiv 1\pmod{3}$, we have \[\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.\]

Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have $S_5\equiv 1\pmod{3}$, and $S_8$, $S_9$, and $S_{10}$ are all divisible by $3$. After $3\cdot 3=9$, we have $S_{17}$, $S_{18}$, and $S_{19}$ all divisible by $3$, as well as $S_{26}, S_{27}, S_{28}$, and $S_{35}$. Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{\mathrm{(B)}\ 197}$. -BorealBear

Solution 2 (bash)

Since we have a wonky function, we start by trying out some small cases and see what happens. If $j$ is $1$ and $k$ is $2$, then there is once case. We have $2$ mod $3$ for this case. If $N$ is $3$, we have $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3$ which is still $2$ mod $3$. If $N$ is $4$, we have to add $1 \cdot 4 + 2 \cdot 4 + 3 \cdot 4$ which is a multiple of $3$, meaning that we are still at $2$ mod $3$. If we try a few more cases, we find that when $N$ is $8$, we get a multiple of $3$. When $N$ is $9$, we are adding $0$ mod $3$, and therefore, we are still at a multiple of $3$.

When $N$ is $10$, then we get $0$ mod $3$ + $10(1+2+3+...+9)$ which is $10$ times a multiple of $3$. Therefore, we have another multiple of $3$. When $N$ is $11$, so we have $2$ mod $3$. So, every time we have $-1$ mod $9$, $0$ mod $9$, and $1$ mod $9$, we always have a multiple of $3$. Think about it: When $N$ is $1$, it will have to be $0 \cdot 1$, so it is a multiple of $3$. Therefore, our numbers are $8, 9, 10, 17, 18, 19, 26, 27, 28, 35$. Adding the numbers up, we get $\boxed{(B) 197}$

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_22&oldid=165414"