PLEASE HELP..PRODUCT OF 3 CONSECUTIVE NUMBERS IS 2184

[PINKY22]

I know the answer is 12 x 13 x 14, but just sort of worked it out by process of elimination, how does a 10 year old show workings? Is there a formula? Have googled but cannot find answer, or one that we understand anyway!

Pick 3 consecutive numbers - calculate answer. If the answer is too low, pick three higher consecutive numbers, or if too high choose 3 lower ones. So you're using the information to make your next guess more selective, until you come to the right answer.

Really!! So I was right really, but surely without a calculator it could take her ages!!

Has it got to be without a calculator? Yes, it would take a while!

Approximate.
9X10x11 = c. 1000, so it's more than that.
19x20x21 = c. 8000, so it's a lot lower than that.
14x15x16 = c. ummmmmm 225 x 15 = 3,375
(obviously you'd try some higher numbers as well at the beginning)

Dilys's husband here - you can express the three consecutive numbers as (n-1) x n x (n+1), which works out as n^3 - n. Then you can guess a value of n whose cube is a bit higher than the target, and see whether it's correct. Seems a bit tough for a ten-year-old, though.

Thanks everyone, I think it's beyond her really! I can't wait till they can use calculators!

Is she Y5/6? She should be using calculators now.

Actually she's only 9 (I thought it must be at least age 10 maths!) and in year 4, she has said they are allowed to start to use calculators next year! I'm going to explain the concept that Dilys's hubby has given because she knows her times tables well so maybe it is not that hard, thinking she could do 10^3 - 10 easily then work her way up. Must be why they made the answer 13! She said she hasn't had it explaned to her at school!

I'm pretty sure they mean them to use approximations to narrow dien the answer - they usually do at this age.

Wouldn't expect them to do anything other than trial and error. And I would let them use a calculator.
Sort of thing you would get on yr 6 SATs calculator paper.

If you divide 2184 into it's factors you get: (do it by dividing by 2 until you can't any longer, then 3, then continue through the primes)
2184 = 2^3 x 3 x 7 x 13
So one of the numbers has to be a multiple of 13, and one of the numbers has to be a multiple of 7.
As 7 x 2 = 14, two of the numbers are likely to be 13 and 14. The third number must be either 12 or 15. If it was a multiple of 5, then the product would end in either 5 or 0, so it can't be 15.
So the numbers are 12, 13, 14
Proof is:
12 = 2 x 2 x 3
13 = 13
14 = 2 x 7
So 12 x 13 x 14 = 2 x 2 x 3 x 13 x 2 x 7
Gathering like terms together you get
2^3 x 3 x 7 x 13 , which is the original number divided into factors.

You could do it by using n(n+1)(n+2) = 2184, but factorising a cubic equation isn't simple. If it was quadratic then it's relatively easy.
Quite fun.

PINKY, my DS1 is also in year 4 and has had a few of these kind of maths questions lately. I think the idea is that they are supposed to estimate, and as Feenie said you kind of narrow it down by working out higher and lower answers until you get the right one if that makes sense!