Difference between revisions of "2017 AIME II Problems/Problem 13"

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==Solution==
 
==Solution==
<math>\boxed{245}</math>
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Considering <math>n \pmod{6}</math>, we have the following formulas:
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Even and a multiple of 3: <math>\frac{n(n-4)}{2} + \frac{n}{3}</math>
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Even and not a multiple of 3: <math>\frac{n(n-2)}{2}</math>
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Odd and a multiple of 3: <math>\frac{n(n-3)}{2} + \frac{n}{3}</math>
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Odd and not a multiple of 3: <math>\frac{n(n-1)}{2}</math>
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Considering the six possibilities <math>n \equiv 0,1,2,3,4,5 \pmod{6}</math> and solving, we find that the only valid solutions are <math>n = 36, 52, 157</math>, from which the answer is <math>36 + 52 + 157 = \boxed{245}</math>.
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=12|num-a=14}}
 
{{AIME box|year=2017|n=II|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:47, 23 March 2017

Problem

For each integer $n\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$.

Solution

Considering $n \pmod{6}$, we have the following formulas:

Even and a multiple of 3: $\frac{n(n-4)}{2} + \frac{n}{3}$ Even and not a multiple of 3: $\frac{n(n-2)}{2}$ Odd and a multiple of 3: $\frac{n(n-3)}{2} + \frac{n}{3}$ Odd and not a multiple of 3: $\frac{n(n-1)}{2}$

Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$, from which the answer is $36 + 52 + 157 = \boxed{245}$.

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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