Question

a, b, cand dare positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.

Answer 1

Given : a, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54

To find  :  (a+b+c+d)·a·b·c·d.

Solution:

a+b+ ab = 76,

Here a & b are interchangeable

and both has to be even number  

if a is odd  , b is odd then ab is odd

=> odd + odd + odd = odd  while 76 is even

if a is odd & b is even   then ab is even

=> odd + even + even = odd

Hence a & b has to be even

a+b+ ab = 76,

=>a (1 + b) = 76 - b

=>  a  = (76 - b)/(b + 1)

b = 2  => a = 74/3    not possible

b = 4 => a =72/5  not possible

b = 6 =>  a = 70/7  = 10

( 6 , 10 ) is one solution

b = 8   =>  a  = 68/9   not possible

only Solution  

6 , 10

c+d+ cd = 54

Same as above c & d has to be even and interchangeable

c = (54 - d)/(d + 1)

d = 2  =>  c = 52/3   not possible

d = 4  =>  c = 50/5  = 10

( 4, 10) is one of the solution

d = 6  => c = 48/7   not possible  and less than 8

Hence only Solution  

4 , 10

a +  b  + c  + d  =  6 + 10 + 4 + 10  = 30

abcd =   6 * 10 * 4  * 10  =  2400

(a+b+c+d)·a·b·c·d.    = 30 * 2400  = 72000

(a+b+c+d)·a·b·c·d.  = 72000

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