Figure this one out...

[Draigun]

On topic - how about I attempt to provide a problem? Simplify the following into one term (it's much easier than it looks, if you know the right method):

x^6 + (6 * x^5 * y) + (15 * x^4 * y^2) + (20 * x^3 * y^3) + (15 * x^2 * y^4) + (6 * x * y^5) + y^6

Quote from Eyes: "x^6+6x^5*y+15x^4*y^2+20x^3*y^3+15x^2*y^4+6xy^5+y^6"

[Darth Crater]

No, it can get a lot simpler. I'm looking for a single term, as opposed to the 7 terms in the original function.

No differential equations, though. It's much simpler.

[[$$$]_Aphelion_]

Math was so easy back then all kids had to do was add up chocolate chips like...

Okay kids If I have 5 chocolate chips in one hand and 2 in the other how much do I have *hand goes up* *teach calls on him/her* yes billy I will have seven now if I ate three how many will I have *kids scream* 4!!!!

That's how math was back then...

[(=DK=)Samonuh]

Math was so easy back then all kids had to do was add up chocolate chips like...

Okay kids If I have 5 chocolate chips in one hand and 2 in the other how much do I have *hand goes up* *teach calls on him/her* yes billy I will have seven now if I ate three how many will I have *kids scream* 4!!!!

That's how math was back then...

Was this an attempt at humor or...? I can't tell...

[Bryant]

Really it was much simpler than that. He joined on your side in exactly the same manner as Marth, he mentioned a few posts up that he's a modder, and I feel like I've heard you use the name "Outrider" in the context of your team before.

On topic - how about I attempt to provide a problem? Simplify the following into one term (it's much easier than it looks, if you know the right method):

x^6 + (6 * x^5 * y) + (15 * x^4 * y^2) + (20 * x^3 * y^3) + (15 * x^2 * y^4) + (6 * x * y^5) + y^6

(x+y)^6

I forget the name of it, but it uses:
121
1331
14641
...

[Outrider]

That would be called a palindrome. ;)

Huh, so it was just a simple reverse binomial expansion. Nice. :o

[kjeopardy]

Damn...people posted the answer before I could attempt Crater's problem...

But yes, (x+y)^6 is correct, and this could be verified with Pascal's triangle and the binomial expansion theorem...

I'll post a problem....this isn't mine, it's from an application to a math program...

Call an integer n revered if:

-Its not divisible by 5.

-Its base-5 expansion is the reverse of its base-8 expansion.

Find all values of n , or show that there are none.

[kjeopardy]

That would be called a palindrome. ;)

Huh, so it was just a simple reverse binomial expansion. Nice. :o

It's actually not a palindrome, since it's not identical from back to front (the exponents are switched). Only the coefficients form a palindrome.

In general, I don't think that there can be any palindrome polynomial functions...think about it: if there are any identical terms, then they would combine (since you always combine like terms) to form a new polynomial, which wouldn't be a palindrome...Hmm...

[Darth Crater]

(x+y)^6

But yes, (x+y)^6 is correct, and this could be verified with Pascal's triangle and the binomial expansion theorem...

Yes, that's correct.

Pi's newest problem...

1 through 4 are revered. Thus, the task is to find all values of n, not show that there are none.

A revered number must have the same number of digits in its base-5 and base-8 forms. This restricts search to the integer ranges 1-4, 8-24, 64-124, 512-624. Limiting the search space further, all base-5 expansions of integers > 512 begin in 4, so we only need to search 1/8 of 512-624.

Brute-force search found only 91.

So, 1, 2, 3, 4, 91, and the negatives of those?

[kjeopardy]

(x+y)^6

But yes, (x+y)^6 is correct, and this could be verified with Pascal's triangle and the binomial expansion theorem...

Yes, that's correct.

Pi's newest problem...

1 through 4 are revered. Thus, the task is to find all values of n, not show that there are none.

A revered number must have the same number of digits in its base-5 and base-8 forms. This restricts search to the integer ranges 1-4, 8-24, 64-124, 512-624. Limiting the search space further, all base-5 expansions of integers > 512 begin in 4, so we only need to search 1/8 of 512-624.

Brute-force search found only 91.

So, 1, 2, 3, 4, 91, and the negatives of those?

You are correct . I've attached a paper solution, although your computer one works in this case. I really only meant the positive integers, so you can disregard the negative ones, although I should have specified that . 91 is the odd one, the others are obvious. If you have any other problems (non-factoring ones, those aren't that much fun ), I'd love to see them. I'll try to find another one...