1. ## Real Numbers

The product of 3 consecutive numbers is 1716. Find their sum.

2. Originally Posted by Drdj
The product of 3 consecutive numbers is 1716. Find their sum.
Let the middle number be $n$ and let the sum of them be $y$

$n(n-1)(n+1) = 1716$

$n(n^2-1) = 1716$

$n^3-n - 1716=0$

Spoiler:
Solve the cubic. As $n^3 > n$ you can get a rough idea by finding the cube root of 1716

$\sqrt[3] {1716} = 11.972$ so guess 12.

$12^3 - 12 - 1716 = 0$

$n = 12$

$y = (n-1) + n + (n+1) = 3n = 36$

3. Thanks so much for the super-swift reply !!
I was kinda stuck at the cubic expression, and how to do a guess-n-check properly - your cube root method was impressive indeed

Just out of curiosity, is there a non-algebraic way to solve it ?

4. I am a bit puzzled as to why you would want a "non-algebraic" way when algebra works so nicely. However, you could try this- since "three consecutive numbers" are about as close togetheras three integers can be, they must each be approximately 1716/3= 572. Try three consecutive numbers around that.

I am wondering why you titled this "real numbers" when, to be able to talk about "consecutive" they must be integer.

and, of course, this has nothing to do with "Statistics and Probability".

5. Originally Posted by Drdj
Thanks so much for the super-swift reply !!
I was kinda stuck at the cubic expression, and how to do a guess-n-check properly - your cube root method was impressive indeed

Just out of curiosity, is there a non-algebraic way to solve it ?
I should point out that I made up that method on the spot. It won't work for small $|n|$ though. You do get a knack for it. For example 10 would be too small

6. Originally Posted by HallsofIvy
I am a bit puzzled as to why you would want a "non-algebraic" way when algebra works so nicely. However, you could try this- since "three consecutive numbers" are about as close togetheras three integers can be, they must each be approximately 1716/3= 572. Try three consecutive numbers around that.

I am wondering why you titled this "real numbers" when, to be able to talk about "consecutive" they must be integer.

and, of course, this has nothing to do with "Statistics and Probability".
Hi,

But since we're talking about the product of 3 consecutive numbers, it shouldn't be 1716/3, but $\sqrt[3]{1716}$

Without the cube root method, one can try to factor 1716 : /3=572, /2=286, /2=143 and 143=11*13.
So 1716 turns out to be 11*12*13

7. Originally Posted by Moo
Hi,

But since we're talking about the product of 3 consecutive numbers, it shouldn't be 1716/3, but $\sqrt[3]{1716}$

Without the cube root method, one can try to factor 1716 : /3=572, /2=286, /2=143 and 143=11*13.
So 1716 turns out to be 11*12*13