OT: Interesting math problem...

[Fyron]

Ragnarok said:
The thing is the book he had didn't indicate if numbers could share the same value. It would be nice if they could as everyone said 00000+00000=00000 is a simple solution.

Unless this book is reknowned for giving trick questions, this is most likely not the solution.

[Ragnarok]

That is what I am thinking. As far as I know the book is just a regular math book for his grade. Unfortunately they cut the answers out of the back...

[Aiken]

Fyron, that's why I quickly deleted my post. But not quickly enough.

I remeber that I saw such tasks in a school. Common way to solve them is to rewrite example to:

10000*T+1000*R+100*I+10*E+D+10000*D+1000*R+100*I+1 0*V+E-10000*R-1000*I-100*V-10*E-T=0

With additional condition it should be quite easy to solve.

[spoon]

Got it:

17465 +
57496 =
-------
74961

T=1
R=7
I=4
E=6
D=5
V=9

I brute forced it, not having time for fancy shmancy formulae. (so, I could be wrong, very wrong!)

[Alneyan]

This problem is the proof I no longer know to do additions; while I figured out the V=9 part, I didn't think that I+I=4 could also mean I=7, and not only I=2.

[geoschmo]

Yay Spoon. I was getting ready to post that it was unsolvable. I had worked out that V was 9 and I was 4. But I got hung up thinking then that the only option for R was 2.

[Ragnarok]

Yay for spoon! Thanks a lot man! I really do appreciate it. Go have a few brewski's at the Cantina on my tab.

[vanbeke]

It looks like is correct
17465 +
57496 =
_____
74961

Brute force solving (sort of)
Assumptions are that all letters represent different numbers and standard notation is used (the first number in a string is not zero)
V must be either 0 (if D+E < 10) or 9 (if D+E > 9)
If V is 0 then I is 5; if V is 9 then I is 4
For either value of V, R must be either 2 or 7; but R can't be 2 since that would require both T and D to be 1
So R is 7
Then T+D = 6 I tried the 4 possibilities and the only one that worked is the form that I gave above.

[Instar]

Its a system of equations, linear algebra might get it done

[Will]

Narf, the equation means you take the constant e (2.71828183...) to the power of one half times the natural logarithm of x, or log base e of x. It doesn't have to be e though, you can get away with something like 10. So, 10^(0.5 * log(x)). It works because of the laws of logarithms, since a^b=c is equivalent to log(a)(c)=b. Then, it is simple to reconstruct as a^(log(a)(c))=c. Square root is simply a number to the one half power, so 10^(0.5 * log(x)) is the sqrt(x).

Either way, you're left with pretty much imprecise methods. I remember they briefly showed how to do it in high school, and that's probably an algorithm like the one Fyron pointed to, and it would be written in an equation as the Riemann sum of varying powers of 10, which would be a very nasty looking thing. Jack's method is easier to write, so I would go with that one. To write it in terms of an equation for square root, it would be something like:

lim(x->&#8734 R(x,N) = sqrt(N)

--edit: ^ that up there is supposed to be the infinity symbol.