# 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,

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

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

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

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

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,

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

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

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

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

$\displaystyle 64(2+3)=4(71+9)$