Re: OT: Interesting math problem...
 

Will, that's a good thourough proof there, but I'm curious why you "discount V=0 as uninteresting". Maybe that's a math thing I don't get, but it seems V=0 is just as valid a posibility as V=9 until it's proven or disproven. In fact I did disprove V=0 in this case, in a very similer fasion to how you proved V=9.

If V=0 then

D + E < 10, since no number for E + 0 + 1 = E or 10 + E
Since D + E < 10, there is no carry over to the second or third columns.
Therefore I = 5, since I must be either 0 or 5 and we have already determined V = 0.
I + I gives us a carry to the next position so R + R + 1 = 5. For this to be correct R must be either 2 or 7. (Somehow I got this point while disproving V = 0, but forgot it while attempting to prove V = 9 ) If R is 2 then there is no carry and T + D is 2. For this to be correct either T and D must both be 1 or T and D are 0 and 2. Both violate our assumptions so R cannot be 2, therefore R = 7.
Therefore 1 + T + D = 7, T + D = 6. T and D cannot both be 3, and neither T or D can be 0(+6) or 5(+1), so T & D must be 2 & 4. To determine which is which we can try and substitute that into D + E = T (Which as we have already calculated is < 10)
If D is 2, E would have to also be 2 for T to be 4, so D cannot be 2.
If D is 4, E would have to be -2 for T to be 2, so D cannot be 4.

Since D cannot be 2 or 4 then it follows that V cannot be 0

Heh, reading this out loud I realized I sound like Vizzini from the Princess Bride. http://forum.shrapnelgames.com/images/smilies/happy.gif

Re: OT: Interesting math problem...
 

Took me about 5 minutes http://forum.shrapnelgames.com/images/smilies/cool.gif
<font class="small">Code:</font><hr /><pre>
9 8 8 9
_______________
1 1 5 | 1 1 3 7 2 3 5
1 0 3 5
========
1 0 2 2
9 2 0
========
1 0 2 3
9 2 0
========
1 0 3 5
1 0 3 5
========
0
</pre><hr />

Re: OT: Interesting math problem...
 

I cannot even understand SJ's question. Am I supposed to be able to comprehend the answer? *Goes back to playing with his abacus*

Re: OT: Interesting math problem...
 

Well, I didn't really elaborate it very well. You have to look ahead a bit to see that V = 0 leads to the trivial (non-distinct) solution of 00000 + 00000 = 00000. While it's not immediately obvious that this is the case, since I don't know of a proof that would be less than your about 9 step proof, there is also a sort of mathematical intuition there, that there already is a trivial solution for zeros, and a non-trivial solution most likely will not have the zeros. But, overall, I was just looking to solve it quickly, elegantly, and I determined that rather than proving V != 0, and doing most of the problem twice, I would just ignore that possibility, and go with the more "interesting" one. And, usually, that works http://forum.shrapnelgames.com/images/smilies/happy.gif

Re: OT: Interesting math problem...
 

Does one of you math geniuses know what the equation for finding the square root of something is?

Re: OT: Interesting math problem...
 

It is rather complicated. This site might prove of some interest. Just found it on Google, but looks interesting... http://www.qnet.fi/abehr/Achim/Calcu...ier_rods2.html

Re: OT: Interesting math problem...
 

That's not an equation, it's a, well, I'm not sure what to call it. A human program in english? My math knowledge says an equation is impossible, by the way, because with an equation you're trying to find a) the number that was multiplied and b) the number it was multiplied by(Please note that while these numbers are equal, for the purpose of reversing the process they must be seperate, even though knowing one means you know the other). Now, we know c), the number to find the square root of. Let's say, 64. The square root of 64 is 8. But, to figure out the equation, we have to forget that except to check to see if we have the right answer.

So, a)=? b)=? and c)=64. Is there an equation that will give a) and b) just from c)?

This equation must also work with c)=16.

Triangulation says no.

Re: OT: Interesting math problem...
 

I came here too late and missed all the fun! http://forum.shrapnelgames.com/images/smilies/wink.gif

To calculate the square root of a number, I believe it has to be done numerically, so you either have to sum an infinite series or use an iterative method that starts with an estimate and gives successively more accurate values with each iteration.

Re: OT: Interesting math problem...
 

Narf: I don't know of any cut and dry equation for finding a square root - however, there is a reasonably simple recursive method for approximating a square root to any precision desired:

R(0, N) = N
R(k, N)) = (N/R(k - 1, N) + R(k - 1, N))/2

DO NOT DO THIS RECURSIVLY - run it as a loop, saving the Last value.
Where N is the original number (constant), and k is a method to control the precision.
A sample: N = 16

Root(0, 16) = 16;
Root(1, 16) = (16/16 + 16)/2 = (1 + 16) / 2 = 8.5
Root(2, 16) = 5.1911764...
Root(3, 16) = 4.1366647...
Root(4, 16) = 4.0022575...
Root(5, 16) = 4.0000006...
Root(6, 16) = 4.0000000...

As you can see, it gets there farily quickly.

Re: OT: Interesting math problem...
 

If you're allowed to use logarithms:

sqrt (x) = e^(0.5 * ln(x))