problem to Brilliant ! 3

Algebra Level 5

How many distinct ordered pairs of positive integers ( a , b ) (a,b) are there satisfying the equation :

1 a + 1 b = 1 105000 ? \frac{1}{a} + \frac{1}{b} = \frac{1}{105000}?


The answer is 567.

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Rushi Wawge by rushi wawge , 20, India

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2 solutions

1 a + 1 b = 1 105000 ⟹ 105000 ∗ ( a + b ) = a b ⟹ ( 105000 − a ) ∗ ( 105000 − b ) = 10500 0 2 . S o a l l f a c t o r s o f 10500 0 2 a r e s o l u t i o n s . S i n c e 10500 0 2 = 2 6 ∗ 3 2 ∗ 5 8 ∗ 7 2 . T h e n u m b e r o f f a c t o r s a r e . ( 6 + 1 ) ( 2 + 1 ) ( 8 + 1 ) ( 2 + 1 ) = 567 , a n d t h i s w o u l d b e t h e n u m b e r o f ‘ p a i r s . \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{105000}\\ \implies~105000*(a+b)=ab\\ \implies~(105000-a)*(105000-b)= 105000^2.\\So ~all~factors~of~105000^2 ~are~solutions.\\Since~105000^2= 2^6*3^2*5^8* 7^2.\\The ~number~of~ factors~are~ .(6+1)(2+1)(8+1)(2+1)= \boxed{ 567 },\\ and~ this~ would~ be ~the~ number~ of`~ pairs.
At least two problems with 50,and 2015 in place of 105000 has already appeared in Brilliant. And this solution was already given by Calvin Lin.

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I love Simon's Favorite Factoring Trick. There are many problems like this, of course, but it's good practice still.

tytan le nguyen - 6 years, 5 months ago
Samanvay Vajpayee
Jul 17, 2015

Using Simon's Favorite Factoring Trick We get, (105000-a)*(105000-b) = 105000^2 For 105000 , we have (8+1)(6+1)(2+1)(2+1)=567 solutions

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