Question

Let p and q be two natural numbers. 

5 p + 3 q + p q = 290.

Calculate q/p.

Easy question.

Answer 1

Given  p q + 5 p + 3 q = 290
       p (q + 5) + 3 q = 290
       p (q + 5) + 3 (q + 5) = 290 + 3 * 5
       (p + 3) (q + 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  q+5 = 5  as q ≠ 0.

  Hence  p + 3 = 5     So  p = 2.
      and   q + 5  = 61    So   q = 56.
       
    q/p = 56/2 = 28.

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From        p (q + 5) + 3 q = 290

we can also do in the following way:

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

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

Since q ≠ 0, q = 56.  Hence, p = 2.

Answer 2

Hello respected sir ^_^!!
This is Aishwary

5p + 3q + pq = 290
adding 15 on both the side , to get

15 + 5p + 3q + pq = 290 +15

5 ( 3+ p) + q( 3+p) = 305

( 5+q) ( 3+p) = 305

( 5+q) ( 3+p) = 5×61. ( As 305 can be expressed in its primes form)
comparing both the side

5 +q = 61
q = 56

and also , 3 +p = 5 => 2

As the question given condition is that
q/p
so
56/2 = 28