Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
japiga
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm
Contact:

Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

Postby japiga » Thu Sep 23, 2010 11:48 am

Find all combinations of the three natural numbers which their sum is equal to their product of those factors (three natural numbers).
I have found just one combination 1 + 2 + 3 = 1 * 2 * 3. But it is only one combination, and the question is: is it only one combination and how to prove it?

User avatar
stapel_eliz
Posts: 1734
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Thu Sep 23, 2010 7:28 pm

japiga wrote:...how to prove it?

Use algebra.

Assuming the numbers are meant to be integers or whole numbers, and assuming that they are meant to be consecutive, then you are trying to find solutions to x + (x + 1) + (x + 2) = x(x + 1)(x + 2).

japiga
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm
Contact:

Re: Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

Postby japiga » Fri Sep 24, 2010 7:47 am

Yes, and I got results (three solutions for x, x1 = 1, x2 = -1 i x3 = 3), and consequently, we have 3 consecutive numbers of each of x with following equations: 1 + 2 + 3 = 1 * 2 * 3; -1 + 0 + 1 = -1 * 0 * 1 and 3 + 4 + 5 is not equal to 3 * 4 * 5 (so in this case we my exclude solution for x3, or to put other numbers in opposite order, such as: 3, 2, 1 and now it gives a sense, 3 + 2 + 1 = 3 * 2 * 1. Do you think that I’ve solved it?

User avatar
stapel_eliz
Posts: 1734
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Fri Sep 24, 2010 1:57 pm

japiga wrote:Yes, and I got results (three solutions for x, x1 = 1, x2 = -1 i x3 = 3)....

How did you arrive at these solutions? (If you plug them back into the original equation, provided earlier, which one(s) work?) :wink:

japiga
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm
Contact:

Re: Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

Postby japiga » Fri Sep 24, 2010 2:18 pm

It works only if x1=1 and x2=-1, but not for x3=3. Is it now correct answer on it?
I arrived on it just solving the equation: x + (x + 1) + (x + 2) = x(x + 1)(x + 2)
x + x + 1 + x + 2 = x(x2 + 2x + x +2)
3x + 3 = x3 + 3x2 + 2x
x3 - 3x2 – x + 3 = 0
x2 (x- 3) – (x – 3) = 0
(x2 – 1)(x -3) = 0

User avatar
stapel_eliz
Posts: 1734
Joined: Mon Dec 08, 2008 4:22 pm
Contact:

Postby stapel_eliz » Fri Sep 24, 2010 7:38 pm

japiga wrote:3x + 3 = x3 + 3x2 + 2x
x3 - 3x2 – x + 3 = 0

You were correct to here:

. . . . .

You subtracted the and the from the left-hand side to the right-hand side:

. . . . .

Then you simplified. But how did you end up with your second line in the quote above?

japiga
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm
Contact:

Re: Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

Postby japiga » Sat Sep 25, 2010 6:14 pm

II have just solved it! We have 3 solutions for x:
x1=1; x2=-1; and x3=-3. So, we have same result when we summarize and multiply all factors x, (x+1) and (x+2).


Return to “Advanced Algebra ("pre-calculus")”