here's someting to get you thinking!?

Question

John, Mary, Thomas and Sheila all have mobile phones with the same company, so all four of them have different numbers. One day, while quite bored, they decide to play the following game. First, each of them subtracts the numbers of the other three people from their own number, then multiplies the three resulting differences together, and finally divides 1 by the result.

So, for instance, if the numbers of the four friends were J (for John's number), M (for Mary's number), T (for Thomas' number), S (for Sheila's number), then John would calculate

1 divided by the product (J - M)x(J - T)x(J - S),

while Mary would calculate

1 divided by the product (M - J)x(M - T)x(M - S).


If John, Mary, Thomas and Sheila add up the four answers obtained in this way, what will the sum be?

Answer 1

Wow! Your math teacher must really not like you to ask this question! =X first I am going to set up the problem for you now you have to assume that none of them have the same number or else you'll be dividing by zero and since we don't know what the sum will be how about we say it's R?

1/(J - M)(J - T)(J - S) +
1/(M - J)(M - T)(M - S) +
1/(T-J)(T-M)(T-S) +
1/(S-J)(S-M)(S-T) +
=R
Now that is pretty ugly so let's make it a little prettier by doing saying
A=(J-M), B=(J-T), C=(J-S), D=(M-T), E=(M-S), F=(T-S)
now we have
1/ABC + 1/(-A)(DE) + 1/(-B)(-D)F + 1/(-C)(-E)(-F) = R
now i'll simplify the negatives and group them together which gives us:
1/ABC + 1/BDF - (1/ADE + 1/CEF) = R
So now what we need to do is use partial fractions on this puppy
so it becomes:
(x1/A+x2/B+x3/C) + (x4/B+x5/D+x6/F) - (x7/A+x8/D+x9/E) - (x10/C+x11x/E+x12x/F) = R
phew! now let's at one case I swear I'm going to have carpal tunnel syndrome after this is done. Let's group some things together I'll use dots to show repetition
(x1-x7)/A + (x2-x4)/B + ... +(x6-x12)/F = R

So let's focus specifically on (x1-x7)/A first find x1
(x1/A+x2/B+x3/C) = 1/ABC mult by A
x1 + x2(A/B) + x3(A/C) = 1/BC now A = J-M so if take the limit when J = M you get A = 0
x1 = 1/(B)(C) @ J=M becomes x1 = 1/(M-T)(M-S) now let's look at x7

(x7/A+x8/D+x9/E) = 1/ADE
x7 + x8(A/D) + x9(A/E) = 1/DE = 1/(M-T)(M-S) again we have to make the A disappear to find x7 so take the limit at J=M and we find quite quickly
x7 = 1/DE = 1/(M-T)(M-S) = x1 !!! so substituting back into our equation
(x1-x7)/A + (x2-x4)/B + ... +(x6-x12)/F = R
(x1-x1)/A + (x2-x4)/B + ... +(x6-x12)/F = R
0/A + (x2-x4)/B + ... +(x6-x12)/F = R
(x2-x4)/B + ... +(x6-x12)/F = R
You then solve for the other x's and you find out that
R = 0
Pretty nifty little equation I think it's actually true for any amount of numbers =)

Answer 2

I'm guessing the answer is 1 or 0. Because the answer to these is always 1 or 0.

Answer 3

cant tell the exact sum as there is not enough info but it would be between 0 and 1

Answer 4

I dont get it and I dont feel like taking the time to figure out the answer

Answer 5

1(?)

Answer 6

u didnt give enough information. u didnt say what the numbers was. how can we figure it out

Answer 7

I do not feel like figuring this out!!!