How many solutions?

How many ordered pairs of positive integers ( a , b ) (a,b) , satisfy the following equation?

a b = 120 ( a + b ) ab = 120\left( {a + b} \right)


The answer is 63.

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2 solutions

Trevor Arashiro
Dec 13, 2014

Rearranging

a b 120 ( a + b ) = 0 ab-120(a+b)=0

a b 120 ( a + b ) + 12 0 2 = 12 0 2 ab-120(a+b)+120^2=120^2

( a 120 ) ( b 120 ) = 12 0 2 (a-120)(b-120)=120^2

Set x = ( a 120 ) x=(a-120) and y = ( b 120 ) y=(b-120)

x y = 12 0 2 xy=120^2

Solving for possible values of x, we see that over the integers, x must be a factor of 12 0 2 120^2 . If we factor 120, we get 120 = 2 3 3 5 120=2^3\cdot 3 \cdot 5

12 0 2 = 2 6 3 2 5 2 \therefore 120^2=2^6\cdot 3^2 \cdot 5^2 .

This has 7 3 3 7\cdot3\cdot3 factors ( explained here )

Therefore, our answer is 63.

Hey can you tell me how did you inserted the link in the phrase "explained here" ?

Vighnesh Raut - 6 years, 5 months ago

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Sure, you put title

Trevor Arashiro - 6 years, 5 months ago
Bill Bell
Apr 16, 2015

b = 120 + 14400 A b=120+\frac { 14400 }{ A } with A = a 120 A=a-120 .

Then A must be one of the divisors of 14400, and there are 63 of those.

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