Integral divisors

Find the sum of all integers n n such that n 2 + 13 n + 2 3 n + 5 \dfrac {n ^{2} +13 n +2} { 3n +5} is an integer.


The answer is 20.

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1 solution

Mathh Mathh
Jun 14, 2014

3 n + 5 n 2 + 13 n + 2 3 n + 5 ( 3 n + 5 ) ( 1 3 n + 34 9 ) 152 9 3n+5\mid n^2+13n+2\iff 3n+5\mid (3n+5)(\frac{1}{3}n+\frac{34}{9})-\frac{152}{9} 3 n + 5 ( 3 n + 5 ) ( 3 n + 34 ) 152 3 n + 5 152 \iff 3n+5\mid (3n+5)(3n+34)-152\iff 3n+5\mid 152

Note that one of the steps I used required gcd ( 3 n + 5 , 9 ) = 1 \gcd(3n+5, 9)=1 , which is true for all n Z n\in\mathbb Z .

Since 3 n + 5 152 3n+5\mid 152 , we have the posibilities 27 -27 , 8 -8 , 3 -3 , 2 -2 , 1 -1 , 1 1 , 11 11 , 49 49 .

Add those up and the answer is: 20 \boxed{20} .

I didn't understand the 2nd step in which you omitted 9 from denominator

VIDIT LOHIA - 6 years, 8 months ago

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I multiplied ( 3 n + 5 ) ( 1 3 n + 34 9 ) 152 9 (3n+5)(\frac{1}{3}n+\frac{34}{9})-\frac{152}{9} by 9 9 .

mathh mathh - 5 years, 7 months ago

Y did u omit 9 from the denominator ?

Bala vidyadharan - 5 years, 7 months ago

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I multiplied ( 3 n + 5 ) ( 1 3 n + 34 9 ) 152 9 (3n+5)(\frac{1}{3}n+\frac{34}{9})-\frac{152}{9} by 9 9 .

mathh mathh - 5 years, 7 months ago

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