[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this cryptarithmetic equation. None of A and B is zero, and C is ≥ 2.

B

∫ x^-C dx = (BDEDFC)/1000000

A

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this cryptarithmetic equation. None of A and B is zero, and C is ≥ 2.

B

∫ x^-C dx = (BDEDFC)/1000000

A

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

Gawd it's been years since i did calculus

(based on the constraints - we don't have to worry about special cases )

The antidirivitive is...

∫ x^-C dx = x^-C+1 (-C+1)^-1

Annndddd....from A to B...

(B^-C+1 (-C+1)^-1) - (A^-C+1 (-C+1)^-1) = (BDEDFC)*10^-6

That's all i remember hahaha Someone else can work on the rest

[Guest]

I have no intention of solving this puzzle, but i just wanted to point out the redundancies in your description. I'm not trying to shoot you down at all, but I figured other people who solve these things might catch this as well.

"C is ≥ 2." is redundant. The ≥ sign implies that C "is greater than or equal to" so the use of "is" in the sentence is unnecessary.

[Guest]

I did this through trial and error with a good amount of educated guessing, but here's what I got and how I got there. I decided it was most probable that A = 1, as the question only specifies that C is above 2, so I decided that (B^-(C-1))/(C-1) - (A^-(C-1))/(C-1) would be negative to ensure that the fraction BDEDFC/(1000000/(C-1)) would be equal to ((B^(C-1))^-1)-1. Those are a lot of liberties I took, but you've got to start with some foundations. I then tried plugging in C as different values, starting at 2 and increasing (I had decided B would be 2 and A would be 1 for all tests in this first round). Luckily enough, when I got to 5 for C, it all worked.

I'm sure this isn't the most professional, but the answers work.

[Glycereine]

I have no intention of solving this puzzle, but i just wanted to point out the redundancies in your description. I'm not trying to shoot you down at all, but I figured other people who solve these things might catch this as well.

"C is ≥ 2." is redundant. The ≥ sign implies that C "is greater than or equal to" so the use of "is" in the sentence is unnecessary.

Although you are correct, your post served no purpose except to "shoot down" the OP. There was no reason or need for it.

Well this has already been solved now but I'd be interested to see if that's the only solution. I'll work on it at lunch.

[Guest]

I got the same answer as the previous post by noting that the result must have exactly six decimal places. This limits the possible values of C, A and B. From there I just worked through the different possible combinations and got that result.

[Guest]

I did find a solution (same as the others here) but it's a bit longer winded than I'd like. I've put the general outline here...

First show that the fracton cannot be in lowest terms, then eliminate teh even values of C. Next show that A and B must have opposite parity and that (AB)^4 must divide 2^6 5^6. The solution is pretty obvious from there.