What is the value of a+b+c+d+e?

[Ryuzaki]

Homework Statement

a, b, c, d and e are distinct integers such that (5-a)(5-b)(5-c)(5-d)(5-e) = 28. What is the value of a+b+c+d+e?

Homework Equations

N/A

The Attempt at a Solution

I tried to solve this in the following manner:- On factorizing 28, we get 28 = 2x2x7.

So, (5-a)(5-b)(5-c)(5-d)(5-e) = 2x2x7.

Now, since a, b, c, d and e are all distinct, each (5 - #) term [where # = a, b, c, d or e] would be a distinct integer.

So, all I need to find are 5 distinct integers whose product gives me 28. The only way I could think of was 28 = 2x(-2)x7x1x(-1).

So, according to the above method, a+b+c+d+e would be 7. But this is not the correct answer, it seems. Can anyone help me on this?

Also, is there a strictly analytical solution to these kind of problems, rather than simply guessing the integers [in this case, it was rather easy to do so!]?

Any help is appreciated. Thank you! :)

 

[jedishrfu]

okay so you got a set of factors 2, -2, 7, 1 and -1 to give you 28

recall the a,b,c,d,e are all part of factors (5-a) ,... so you still need to do some simple math to get a,b,c,d,e and then add them up.

analytically, you can set limits on the factors such as no factor can be 0 and you can demonstrate that no factor can be greater than 28 or actually 14 and that at least two must be negative...

 

[Ryuzaki]

Oh, that was rather silly of me!

So, a, b, c, d and e would be actually 3, 7 , (-2), 4 and 6. And thus, a+b+c+d+e would be 18.

Thank you, jedishrfu! :)