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.

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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.

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Amelia 4 weeks 2021-12-26T10:37:58+00:00 1 Answer 0 views 0

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    2021-12-26T10:39:55+00:00

    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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