Question

what snappy reasoning shows that
\[3^{100}-2^{100}\] is divisible by 11.

Answer 1

i found a solution by factoring but it was not at all obvious, so i hope there is a simple 'splanation

Answer 2

U mean apart from your amazing feat of mental calculation just recently:-)

Answer 3

well i factored.
what mistake did i make now?

Answer 4

Just kidding...

Answer 5

i factored
\[x^{100}-y^{100}\] and one of them, when you make the replacements, gives 55. but i thought there might be some simpler snappier number theoretic ish reason

Answer 6

Doing one at a time is no problem, I'm just mulling over the subtraction.

Answer 7

maybe there is a snappy way to factor.
you get
\[(x-y)(x+y)( \text{ this one gives 55})(\text{other stuff})\]

Answer 8

Maybe u can do it:

3^100 = (3^10)^10 = 1 mod 11 and
2^100 = (2^10)^10 = 1 mod 11 by FLT and I see no reason not to subtract these in terms of the remainders, 1-1 = 0 mod 11

Answer 9

There is only one factor divisible by 11. Refer to the attached Mathematica calculation.

Answer 10

The number theory method isn't concerned with the factors only the remainders.

Answer 11

thanks! i was fairly certain there was a snap way to do this. thanks!