Question

Show that 2^99 +3^99 is divisible by 35.

Answer 1

i have an inelegant method, but i am sure there is a snappier way
you can factor
\[x^{99}+y^{99}\] as the sum of two cubes, and continue on down
when you get done one factor will be \(x+y\) which in your case is \(2+3=5\)
the next factor will be \(x^2-xy+y^2\) which gives you \(7\)

hello joe!
now comes a better answer

Answer 2

If you dont use the standard modular arithmetic way, you could write the problem as:\[(2^3)^{33}+(3^3)^{33}\]and factor it.

Answer 3

yo Satellite :)

Answer 4

Can you please show me how to factor the expression?

Answer 5

i guess we are thinking alike
nice to see you again! on break?

Answer 6

\[x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots +(-1)^n y^{n-1})\]if n is odd. I might be off on the sign of the last term, but it doesnt matter for this problem.

Answer 7

that gives the 5, you need the next term for the 7

Answer 8

yeah, sign is wrong, that last term should be (-1)^n-1 y^n-1

Answer 9

if you factor with the 33 as the major power instead of 99, you end up with the whole 35.

Answer 10

maybe there is a better way. what class is this for?

Answer 11

\[(2^3)^{33}+(3^3)^{33}=(2^3+3^3)(junk)\]

Answer 12

much nicer

Answer 13

that makes perfect sense, i did a similar question before but with difference of powers instead of the sum. Now I know that this method works for both situation. Thanks! This is math 9 and i am doing factoring difference of squares and cubes.

Answer 14

\[2^{99}+3^{99}=(2^3)^{33}+(3^3)^{33}=8^{33}+27^{33}=(8+27)(8^{32}-8^{31}27+\cdots+27^{32})\]

Answer 15

yep im on break sate, and i needed it too >.> last semester was brutal =/ only one more semester till I get my bachelors though :) im applying to grad schools now, its a pain >.<

Answer 16

Thank you so much for helping!!

Answer 17

they should just give you the masters now
then you can spend the next 7 years on phd