Question

a and b are two non-negative integers. Given

$1 + \frac{5}{a} + \frac{3}{b} = \frac{290}{a \: b}$
Find a b and a/b ...

Answer 1

Given   1  + 5/a  +  3/b  =  290/(a b)

Multiply with a b on both sides.  We know a ≠0 and b ≠ 0 .

So  a b + 5 b + 3 a = 290
       b (a + 5) + 3 a = 290
       b (a + 5) + 3 (a+5) = 290 + 3 * 5
       (b + 3) (a + 5) = 305 = 61 * 5

     61 and 5 are prime numbers.  So we can say that one of the factors on LHS is equal to one of the factors on the RHS.

      We cannot say  a+5 = 5  as a ≠ 0.

  Hence  b + 3 = 5     So  b = 2.
      and   a + 5  = 61    So   a = 56.
       
    a/b = 56/2 = 28.

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From        b (a + 5) + 3 a = 290

we can also do in the following way:

$b = \frac{290 - 3 a}{a+5}=\frac{290-3(a+5)+15}{a+5}\\\\b=\frac{305}{a+5}-3=\frac{61*5}{a+5}$

As b and -3 are integers, 61 * 5 must be  multiple of a+5.
Since 61 and 5 are prime numbers,  a+5 must be equal to 5 or 61.  

Since a ≠ 0, a = 56.  Hence, b = 2.

Answer 2

Helloooo Sir,

$1 \: + \: \frac{5}{a} \: + \frac{3}{b} = \: \frac{290}{ab}$

= (ab + 5b + 3a)/ab = 290/ab

= ab + 5b + 3a = 290

= b(a + 5) + 3a = 290

Now, add 15 on both sides.

So,

b ( a + 5) + 3(a + 5) = 290 + 15

= (a + 5) ( b + 3) = 305

= (a + 5) ( b + 3) = 61 × 5

Now, Since a cannot be 0

So,

a + 5 = 61
=> a = 56....... (1)

Similarly,

b + 3 = 5
=> b = 2 .......... (2)

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We need to find a/b

=> 56 / 2

=> ●● 28 ●●.......ANS.

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Thanks for asking such questions.

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