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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- May 24th 2007, 10:05 PMinolas_culpls solve
IF THE PRODUCT OF 4 natural nos. is equal to their sum.wat shud b the min value of the sum??

- May 24th 2007, 10:08 PMJhevon
- May 25th 2007, 04:27 AMtopsquark
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