Regular Polygons and 359°
Find the total number of possible n values in which the interior angle of a regular polygon with n sides and the interior angle of a regular polygon with m sides add up to exactly 3 5 9 ° .
The answer is 105.
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3 solutions
Great solution! Thanks for sharing.
Am I missing something? Wouldn't m be negative for all values of n ?
Also, n can not be 1 , 2 (there are no regular polygons with these many sides)
Can you please give an example of ( n , m ) ?
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I did not say that n had to be a positive factor of 3 6 0 2 . I said that n − 3 6 0 had to be a positive factor of 3 6 0 2 . Thus we could have n − 3 6 0 = 1 , m − 3 6 0 = 3 6 0 2 , so that n = 3 6 1 and m = 3 6 0 × 3 6 1 , for example.
We should remove 4 solutions.
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You should read my reply to the previous comment before making the same mistake as Mr India...
1 8 0 − n 3 6 0 + 1 8 0 − m 3 6 0 = 3 5 9 ⟹ n 3 6 0 + m 3 6 0 = 1
We are hence looking for n 3 6 0 = b a for coprime ( a , b ) , and m 3 6 0 = b b − a .
In which case, n = a 3 6 0 b , m = b − a 3 6 0 b . So a and b − a are coprime factors of 3 6 0 .
3 6 0 = 2 3 × 3 2 × 5 .
Possible coprime pairings of factors are:
2 , 3 , 5 ∤ n , Leads to 4 × 3 × 2 = 2 4 pairs.
2 ∣ n , & 3 , 5 ∤ n , Leads to 3 × ( 3 × 2 ) = 1 8 pairs.
3 ∣ n , & 2 , 5 ∤ n , Leads to 2 × ( 4 × 2 ) = 1 6 pairs.
5 ∣ n , & 2 , 3 ∤ n , Leads to 1 × ( 4 × 3 ) = 1 2 pairs.
2 , 3 ∣ n , & 5 ∤ n , Leads to ( 3 × 2 ) × 2 = 1 2 pairs.
2 , 5 ∣ n , & 3 ∤ n , Leads to ( 3 × 1 ) × 3 = 9 pairs.
3 , 5 ∣ n , & 2 ∤ n , Leads to ( 2 × 1 ) × 4 = 8 pairs.
2 , 3 , 5 ∣ n , Leads to ( 3 × 2 × 1 ) × 1 = 6 pairs.
The total number of pairs are 1 0 5 .
Note, for coprime pair of factors ( x , y ) , n = x 3 6 0 ( x + y ) , m = y 3 6 0 ( x + y ) , which are integers, and n 3 6 0 + m 3 6 0 = x + y x + x + y y = 1 .
Nice solution!
Length of table is 105 (the result of { m , n } /. Solve [ m 3 6 0 + n 3 6 0 = 1 ∧ m ≥ 3 ∧ n ≥ 3 , Z ] :
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We want to solve the Diophantine equation 1 8 0 − n 3 6 0 + 1 8 0 − m 3 6 0 m 3 6 0 + n 3 6 0 ( m − 3 6 0 ) ( n − 3 6 0 ) = 3 5 9 = 1 = 3 6 0 2 = 2 6 × 3 4 × 5 2 Thus possible values of n − 3 6 0 are positive integer factors of 3 6 0 2 . There are 7 × 5 × 3 = 1 0 5 such factors.