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

(Created page with "==Problem== For each integer <math> n\geq 2 </math>, let <math> S_n </math> be the sum of all products <math> jk </math>, where <math> j </math> and <math> k </math> are integ...")
 
(→‎Solution 1)
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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
 
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
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==Solution 2 (bash)==
  
 
==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 11:20, 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)

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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