1

Each interior angle of a regular polygon with $n$ sides is $\frac{3}{4}$ of each interior angle of a second regular polygon with $m$ sides.

How many pairs of positive integers $n$ and $m$ are there for which this statement is true?

$\frac{(n-2)*180}{n}$ is the value of one interior angle for a polygon with $n$ sides.

Therefore $\frac{(n-2)*180}{n} =\frac{3(m-2)*180}{4m}$ and $mn - 8m + 6n =0$

For how many pairs of positive integers $n$ and $m$ is the statement $mn - 8m + 6n =0$ true?

Hendrix
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Eris Tyenns
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4 Answers4

4

Hint

$$mn-8m+6n=0\iff (m+6)(n-8)=-48$$

Mostafa Ayaz
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  • So the only pairs I can make with positive integers n and m are $24(-2)$, $12(-4)$, $8(-6)$, $16(-3)$ and $6*(-8)$. I found the pairs (m,n):$ (18,6), (6,4), (2,2), (10,5)$ and $(0,0)$. Should I include $(0,0)$? – Eris Tyenns Aug 20 '19 at 22:16
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    If you are restricted to positive integer $m,n$, the case $6*(-8)$ drops out... – Mostafa Ayaz Aug 20 '19 at 22:18
  • Thanks, but why can't I set m=$0$ and n=$0$? Is $0$ not a positive integer? – Eris Tyenns Aug 20 '19 at 22:21
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    You're welcome. No. $0$ is not positive. Check out https://en.wikipedia.org/wiki/Sign_(mathematics) – Mostafa Ayaz Aug 20 '19 at 22:22
2

$$mn-8m+6n=0$$

$$m(n-8)=-6n$$ $$m=\frac {6n}{8-n}$$ The positive integral solutions are $$(m,n)=(2,2),(6,4),(18,6),(42,7),(10,5) $$

Mohammad Riazi-Kermani
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1

$$mn-8m+6n=0\implies n(m+6)=8m\implies n=\frac{8m}{m+6}$$ Testing this formula from $1\le m\le 1000$, the only integer solutions found were $$\frac{2\times8}{2+6}=2 \quad\frac{6\times8}{6+6}=4 \quad\frac{10\times8}{10+6}=5 \quad\frac{18\times8}{18+6}=6 \quad\frac{42\times8}{42+6}=7$$

If we solve the other way around, $$m=\frac{6n}{8-n}$$ and we can see a limit where $1\le n\le7$ so the pairs are $(2,2), (6,4), (10,5), (18,6), (42,7)$.

poetasis
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0

If you solve for $m$ in terms of $n$ and impose the condition $m>0,$ you discover that $n<8.$ If you now use the fact that $$\frac{6n}{8-n}$$ is an integer, you should be able to manually list all the solutions.

Check: There are five solutions.

Allawonder
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