Anyone able to solve this exam question?

[bartonian]

Three whole numbers when added together make 20. When multiplied together they make 300. What are the numbers?

I am hoping there's a misprint as I can't see how it could be done

It is a Wycombe Abbey Scholarship paper. Photocopy so don’t think it can be a mis-print. We thought maybe it should have been 21 rather than 20 but surely that kind of mistake wouldn’t slip through? I have googled and no answers come up. I am tempted to email the maths teacher and demand the answer!

Is this more about proving that the smallest sum of three factors of three hundred is 5+6+10=21? hence finding three which sum to 20 is impossible?

Give me a little time and I could prove that.

So a paper that should be taken by 11 year olds? Yeah, got to be a mistake!

noblegiraffe. Accepting this is probably a mistake, I don't think this is solvable, but can that be proved?

I realise this is a bit cheeky, but as it's a long time since I ever studied any maths I would relish the chance to understand something challenging like this.

Not a proof, but to explain:
The prime factors of 300 are 2 x 2 x 3 x 5 x 5 so any numbers that multiply to 300 have to be made up of combinations of those numbers multiplied together. E.g. 300 = 10 x 30 which is 2 x 5 =10 multiplied by 2 x 3 x 5 = 30.

We need 3 numbers that multiply to 300 and we have 5 prime factors so we need to pair up some of factors to make other numbers. But they can't add to more than 20 or we go bust.

2 x 2 x 3 x 5 x 5 - pick pairs out of these

So we can't have 5 x 5 as 25 is bigger than 20
5 x 3 is 15 leaving us with 2 x 2 x 5 to take a pair and a single out of, all of which take us over 20 (e.g. 15 x 4 x 5 adds to 24)
5 x 2 is 10 leaving us with 2 x 3 x 5 (10 + 2 + 15 or 10 + 6 + 5 both bust)
If we decide to leave both 5s as singles we have 5 + 5 + 2 x 2 x 3 = 22.

So we can't pair off the 5s with any other factor and we can't have them on their own, so we can't get a sum under 20.

I think that's right - please point out any errors, I was making dinner and typing

That's assuming no negatives, obviously, that might take a bit longer!

Thanks!

Allowing negatives doesn't take much longer. We need two negative numbers, which means we need the first number to be bigger than 20 and the product of two or three of the factors.

553 = 75 (-2-2 = 71)
552 = 50 (-2-3 = 45)
55 = 25 (-2 - (23) = 17)
55 = 25 (-3 - (22) = 18)
532 = 30 (-5-2 = 23)

None of the other products are bigger than 25. It's proof by exhaustion.

Oops, that isn't complete because I forgot we can use -1 as well so we can have four or five numbers in the first term. So it does take a bit longer to write them all out.

55322 = 300 (-1-1 = 298)
5532 = 150 (-1-2 = 147)
5522 = 100 (-1-3 = 96)
532*2 = 60 (-1-5 = 54)