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Maths problem

Here's one for all you mathematicians out there.
(This was originally posed on the BBC Radio 4 forums by Saibal Mitra, so I take no credit for it - and, incidentally, I haven't yet managed to solve it).

The sum of 3 numbers is 4
The sum of the squares of those 3 numbers is 10
The sum of the cubes of those 3 numbers is 22
What is the sum of the 4th powers of those 3 numbers?
Gagnar The Unruly
Scroll and highlight the box below for the answer:


a = 0
b = 1
c = 3

the answer is 64

I solved this by simple reasoning.

1) the numbers would be integers
2) the numbers had to be between -3 and 3, or else the sum of their squares would exceed 10

From here it's easy to test the possible combinations, to make sure that the sum of the squares equals ten. Testing the criterion for the sum of the squares is simplest because you don't have to use negative numbers. The sum of the squares of 1, 2, and 2 = 9 (too low), 2, 2, and 2 = 12 (too high). That means that 3 has to be used as one of the numbers, and it obviously only works with 1 and 0 (because 3^2 = 9).

The answer is simple, if you think to use 0!

Sorry - wrong.
That cubes to 28, not 22.

You didn't really think it would be *that* easy did you?

The second part of your reasoning is sound, but the first isn't.

Hint1: the numbers do not need to be integers and none of them is 0.

Hint2:you can do it by 'brute force' methods or you can look at cubic polynomials...

PS - It DOES have a real solution (ie you don't need imaginary numbers).
Gagnar The Unruly
D'oh! I should've known it wasn't that easy. Don't take it personally, Bikerman. I'll give it a shot, but my brain power and patience might not hold up. Incidentally, I started at the brute force approach and quickly decided it wasn't for me. I'll think harder and see if I can find another way around.

Ironically, I made the same calculation mistake (3*3 = 21?) when buying wood chips for a flower bed. This is why I'm a biologist and not a mathematician.
Ok lets start!


x2 +y2+z2= 10

x3 +y3+z3=22


I think it is some sort of progression series... I can reply later
You don't necessarily have to solve for x, y, and z (though it is possible to do so).
I simply want you to solve for x^4+y^4+z^4=?
Hi Biker

Thanks for such a cool problem! It took me a few hours to solve.

The solution is 50, and the 3 numbers are:


If you're wondering how I found these numbers, I played around bit and tried 0, 1, 3, which give the right sum and square, but with sum of the cubes 28 instead of 22.

I then used partial derivatives to improve the solution, and got -2/3, 2 and 2 2/3.

I then decided to adjust the 1st and 3rd numbers, keeping the 2 fixed (this was the crucial step, and a bit lucky!)

If we assume one number is 2, another x, the last must be 2-x (to sum to 4).

To get the sum of the squares right, we must have x^2 + 2^2 + (2-x)^2 = 10
This simplifies to x^2 - 2x - 1 =0
which has solutions: 1-√2 & 1+√2
which we put together with 2 to get the numbers above, and bingo! the sum of the cubes is correct.

I confess I calculated the sum of the fourth powers using a calculator, but it could be done properly using the binomial expansion

So there you are...

Well done. Correct solution.
(I cannot claim the credit for the problem - it was posed on the BBC Radio 4 boards).

The 'simple' solution is provided by CountIbis over at the physics forums:

PS - I didn't offer a prize but I feel that a donation of FRIH$250 would be in order - consider it done.
This problem was interesting. I spent several hours today working on it. It was frustrating because it can be solved by simple math, but it's not really a simple problem. This reminds me of a math problem that I did last year that was titled "The World's Hardest Easy Geometry Problem" because the entire problem can be solved using simple geometry, but the problem is extremely difficult to figure out. I spent several weeks on that one. I'll provide the link if anyone is interested.
Hi Biker

You're welcome.

I'll certainly take a look at the solution on the link.

And thanks for the Frih$!

BTW - where abouts in Cheshire do you live? I'm from Northenden in South Manchester (used to be in Cheshire)
infinisa wrote:
Hi Biker

You're welcome.

I'll certainly take a look at the solution on the link.

And thanks for the Frih$!

BTW - where abouts in Cheshire do you live? I'm from Northenden in South Manchester (used to be in Cheshire)

I live in a town called Frodsham - between Warrington and Chester..
Hi Bikerman

I checked out the solution by CountIbis over at the physics forums:

It's really clever and exotic. The maths is pretty basic, but there are so many tricks I would never have thought of this...

I notice that physicsforums have LaTeX enabled. It would be really cool to have LaTeX supported on Frihost. Do you think this is possible?

For those who are curious about LaTeX, take a look at the post by rock.freak667 in the link above, and click on the line , and you'll see:
with a link to their LaTex reference .

Any chance?
Well, I will certainly raise it with the system admins - I agree it would be marvellous to have LaTeX enabled on the forums but I'm not sure how easy that would be....I'll get back to you on this one...
If anybody wants to see an algebraic solution without guessing a, b, c or using weird logarithms and stuff, there is one below. Pretty quick to get if you can write fast and expand polynomials in your head:

(a2 + b2 + c2)^2 = a4 + b4 + c4 + 2(a2b2 + a2c2 + b2c2). Need a2b2 + a2c2 + b2c2!

ab + ac + bc = .5 * ( (a+b+c)^2 - (a2 + b2 + c2) ) = .5*(4^2-10) = 3

Square of that is 9 = a2b2 + a2c2 + b2c2 + 2aabc+2abbc+2accb. Need aabc+abbc+accb = abc(a + b + c)! Need abc!

To find abc, consider (a+b+c)^3 - (a^3 + b^3 + c^3) - 3(a + b + c)(ab + ac + bc)  = -3abc, so -3abc = 4^3 - 22 - 3(4)(3), so abc = -2.

This lets us find aabc+abbc+accb = -2(4) = -8. In turn, we can find a2b2 + a2c2 + b2c2 = 9-(2)(-8) = 25.

And with that a4 + b4 + c4 finally can be found to be 10^2 - 2(25) = 50.

Yeah Tex would be pretty nice. It would probably be pretty easy to install, too (admin could just google tex on phpbb).
Well, I have raised the issue of LaTeX with the powers that be. I'll keep you informed of developments/decisions.
Hi nilsmo

That is a totally cool solution, easily the best so far!

It would be perfect, if, as a bonus, you could figure out the actual values of a, b & c.

I hope I'me wrong, but I feel that may be a little harder.
That's ingenious! I'll pose it to my maths class and see if anyone can solve it!
That's an intriguing little maths problem, Bikerman. Thanks for posting it - should be quite interesting to try to solve. Better yet - it should be interesting to see if any of my friends can solve it...
This is fun!!
Thanks I'll share this with my son who seems to be a budding mathematician.
wow thats so cool! i like this forum
I thought it would be easy when I tried to solve it and now.......not so much. Lol
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