I give you a bigrade:

Now give me ab + ac + bc = de + df + ef (please show all work) and find values for a - f that makes the equation work.

Results 1 to 6 of 6

- Dec 28th 2009, 10:52 AM #1

- Joined
- Oct 2009
- Posts
- 769

- Dec 28th 2009, 10:26 PM #2

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Dec 28th 2009, 11:45 PM #3

- Joined
- Jan 2009
- Posts
- 715

- Dec 29th 2009, 10:10 AM #4

- Joined
- Oct 2009
- Posts
- 769

## Response to Captain Black

Of course I know how to solve this problem, otherwise I wouldn't have posted it here (I didn't want to immediately post the solution as that would have given away the show).

simplependulum got the answer right. I would have squared both sides and then use simple algebra to derive the end result to show that it follows naturally - the same difference, but I think my way would be more understandable. Now for the next part (which I know how to do):

Starting from:

, can you show through adding 1 to each and every member of this bigrade that you still have equality? [the first step is to derive ]

- Dec 29th 2009, 10:24 AM #5

- Joined
- Oct 2009
- Posts
- 769

## BTW

Here are one set of values (among the infinite) that make the bigrade work: a = 1, b = 6, c = 8, d = 2, e = 4 and f = 9. The reason why it's important to specify is you have can doubters (like Kronecker) who say that nothing is proven unless you can show everything.

- Jan 3rd 2010, 01:00 PM #6

- Joined
- Oct 2009
- Posts
- 769

## Solving the puzzle

I left it to be proven that:

reduces to:

Opening parentheses, you get:

Now rearranging terms and combining some:

Obviously the 3 cancels out. Also, by assumption, since this is a bigrade,

a + b + c = d + e + f also cancels out from the equation (with the coefficient 2) which leaves you with:

which is true by assumption. Here's my main point: with any multigrade, you can add any positive integer (to each and every term) and get a new multigrade, no exceptions.