- 25 Oct '07 17:14 / 1 editI have 3 numbers, lets call them a, b and c. I am surprised to see that one of the pairs is such that the difference between the two numbers is exactly equal to the difference between their squares.

ie a-b = a^2-b^2

when I look at the other pairs I find that in one pair the difference between their squares is twice their difference and in the final pair the difference between their squares is thrice their difference.

What is the sum of a, b and c?

(The problem is easy ... your solution must be concise!) - 25 Oct '07 18:26

At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.*Originally posted by wolfgang59***I have 3 numbers, lets call them a, b and c. I am surprised to see that one of the pairs is such that the difference between the two numbers is exactly equal to the difference between their squares.**

ie a-b = a^2-b^2

when I look at the other pairs I find that in one pair the difference between their squares is twice their difference and in the final ...[text shortened]... e.

What is the sum of a, b and c?

(The problem is easy ... your solution must be concise!) - 26 Oct '07 08:37

Yes. The solution is unique and a+b+c=3*Originally posted by PBE6***At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.**

But how did you get there? Its the technique I was after!!!

What if the ratios in question are; 1, 49.5 & 149.5 ?????

Show your working for the marks!!

- 26 Oct '07 09:05 / 2 edits

a^2 - b^2 = a - b*Originally posted by wolfgang59***But how did you get there? Its the technique I was after!!**

b^2 - c^2 = 2(b - c)

c^2 - a^2 = 3(c - a)

Note that a^2 - b^2 = (a - b)(a + b)

If a, b and c are all different, we can divide each line by (a - b), (b - c) and (c - a) respectively.

a + b = 1

b + c = 2

c + a = 3

Add them up (and divide by 2): a + b + c = 3.

If the ratios are p, q and r, then the same process shows that:

a + b + c = (p + q + r)/2

If you allow a, b and c to be equal, then you always have (0, 0, 0) as a possible solution. - 26 Oct '07 09:18

Note that a^2 - b^2 = (a - b)(a + b)*Originally posted by mtthw***a^2 - b^2 = a - b**

b^2 - c^2 = 2(b - c)

c^2 - a^2 = 3(c - a)

Note that a^2 - b^2 = (a - b)(a + b)

If a, b and c are all different, we can divide each line by (a - b), (b - c) and (c - a) respectively.

a + b = 1

b + c = 2

c + a = 3

Add them up (and divide by 2): a + b + c = 3.

If the ratios are p, q and r, then the same process shows that:

a + b + c = (p + q + r)/2

Yes!

If you dont spot this and instead solve the simultaneous equations you can get in a mess .. particularly for the second scenario!!