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.
Answers
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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