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  1. #1
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    pls solve

    IF THE PRODUCT OF 4 natural nos. is equal to their sum.wat shud b the min value of the sum??
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by inolas_cul View Post
    IF THE PRODUCT OF 4 natural nos. is equal to their product.
    this statement is redundant, i believe you mistyped the question
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by inolas_cul View Post
    IF THE PRODUCT OF 4 natural nos. is equal to their sum what should be the min value of the sum??
    a + b + c + d = abcd

    Thus
    a + b + c = abcd - d

    a + b + c = (abc - 1)d

    d = (a + b + c)/(abc - 1)

    Thus a + b + c must be divisible by abc - 1.

    I am having a disagreeably difficult time meeting (even envisioning) this restriction unless
    a + b + c = abc - 1

    So
    a + b + 1 = abc - c

    a + b + 1 = (ab - 1)c

    c = (a + b + 1)/(ab -1)

    So a + b + 1 must be divisible by ab - 1.

    I am going to again try
    a + b + 1 = ab - 1

    a + 2 = ab - b

    a + 2 = (a - 1)b

    b = (a + 2)/(a - 1)

    So a + 2 is divisible by a - 1.

    Now a + 2 is never equal to a - 1. So let
    a + 2 = n(a - 1)
    where n is also a natural number.

    Then
    a + 2 = na - n

    a - na = - 2 - n

    na - a = n + 2

    (n - 1)a = n + 2

    a = (n + 2)/(n - 1)

    a must be a natural number, so let's try this for various values of n > 1:
    n = 2 ==> a = 4
    n = 3 ==> a = 5/2
    n = 4 ==> a = 2
    n = 5 ==> a = 7/4
    n = 6 ==> a = 8/5
    etc.

    So it looks like a = 4 or a = 2.

    Use a = 2.
    Then
    b = (a + 2)/(a - 1) = 4
    c = (a + b + 1)/(ab -1) = 7/7 = 1
    d = (a + b + c)/(abc - 1) = 7/7 = 1

    So a = 2, b = 4, c = 1, d = 1.
    (Check: 2 + 4 + 1 + 1 = 8 and 2*4*1*1 = 8.)

    Using a = 4 we get b = 2, c = 1, d = 1 so this gives nothing new.

    I can't find a way to get another set, though from the wording of the question more than one exists.

    -Dan
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