OT: Interesting math problem...

[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.

[Will]

A more rigorous "proof", took me about five minutes before I went to class, posting it now after class:

Code:

---

 TRIED

+DRIVE

------

 RIVET



Assuming we disregard the trivial solution (all zeros), and 

force each letter to be distinct (ie for any x and y in 

{T,R,I,E,D,V}, the values x != y), and noticing that the 

carry can be at most 1 (9 + 9 + 1 = 19):



either E + V = E (V = 0)

    or E + V + 1 = E (V = 9)

disregard V = 0 as uninteresting, so V = 9

    and D + E > 10 and T < D, T < E

either I + I = V = 9

    or I + I + 1 = V = 9

I + I != 9, so I + I + 1 = V is true

so I = {4,9}, but V = 9, so I = 4

so R + R = I = 4 (no carry from previous digit)

then R = {2,7}

V = 9, I = 4, R = {2,7}

either T + D = R = 2

    or T + D + 1 = R = 7

assume there is no final carry

so if T + D = 2, then T = D = 1

then T + D + 1 = 7 = R

then T < (7 - D)

V = 9, I = 4, R = 7, T < (7 - D), T < E

try T = 1:

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

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[Suicide Junkie]

Here is a tough one:
Fill in the long division using the blanks and the given numbers.
Code:

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              _ _ 8 _

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_ _ _ | _ _ _ _ _ _ 5 

        _ _ _ _

        ========

          _ _ _ _

            _ _ _ 

          ========

            _ _ _ _

              _ _ _

            ========

              _ _ _ _

              _ _ _ _

              ========

                    0

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[geoschmo]

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.

[douglas]

Suicide Junkie said:
Here is a tough one:
Fill in the long division using the blanks and the given numbers.
Code:

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              _ _ 8 _

       _______________

_ _ _ | _ _ _ _ _ _ 5 

        _ _ _ _

        ========

          _ _ _ _

            _ _ _ 

          ========

            _ _ _ _

              _ _ _

            ========

              _ _ _ _

              _ _ _ _

              ========

                    0

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Took me about 5 minutes
Code:

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              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

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