# Thread: Puzzle like algebra question

1. ## Puzzle like algebra question

Hey All,

This one feels a like difficult math puzzle. I have no idea where to start on this one. I tried guessing around on the calculator, but that won't really explain the right approach for such a problem, which is what I am looking for.

Given the following 7 numbers,

$2, 3, 4, 9, 64, 71, 91$

Find six numbers a, b, c, d, e and f from the list such that,

$a(b + c) = d(e + f)$

The only interesting thing I see here is the numbers $2, 2^{2} = 4, 4^{3} = 64$. But no clue as to how this is useful.

Thanks for all your help.

2. Originally Posted by mathguy80
Hey All,

This one feels a like difficult math puzzle. I have no idea where to start on this one. I tried guessing around on the calculator, but that won't really explain the right approach for such a problem, which is what I am looking for.

Given the following 7 numbers,

$2, 3, 4, 9, 64, 71, 91$

Find six numbers a, b, c, d, e and f from the list such that,

$a(b + c) = d(e + f)$

The only interesting thing I see here is the numbers $2, 2^{2} = 4, 4^{3} = 64$. But no clue as to how this is useful.

Thanks for all your help.
$64(2+3)=4(71+9)$

3. 64(3 + 2) = 4(71 + 9) ; plus a few rearrangements, like 64(2 + 3)...

4. Thanks, @earboth and @Wilmer. Turns out paper was better option to figure out the right combination.