i have the numbers 3,4,5,6,7,8,28,30,35 i need to put them in to 3 groups and i need to times them together so all three groups =the same
Its actually not hard at all.
If you reduce each number to their prime factorization you will see that you are forced into certain combinations.
For example, each one of the product triplets must have a factor of both 7 and 5.
Furthermore if you break the numbers down to the smallest three, middle three, and biggest three, you'll convince yourself that the product triplets must contain one from each group
Anyway, here is the answer:
They all equal 840.
Actually after giving this a little more thought, there is another clever way to go about this problem. You could multiply ALL the numbers together and get 592,704,000
Now since all three triplets must multiply to the same number this implies that the product of all numbers if a perfect cube, and indeed the cubed root of 592,704,000 is 840. Then you can just use trial and error to find some combination of 3 numbers that multiply to 840, once you have one, a second won't be hard to find, and the numbers left over after that will be your last group.
Neat little problem. Thanks.