Question

Find the smallest positive integer for which\[5^3|2^n+3^n\ .\]

Answer 1

\(n\) must be odd so \[2^n+3^n=5(2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1})\]

Answer 2

Can u PLZ explain the question

Answer 3

@mukushla

Answer 4

@sauravshakya

for n=1 2^1+3^1=5 is divisible by 5 ... i want u to find smallest positive integer n such that 2^n+3^n is divisible by 125

Answer 5

thanx @mukushla

Answer 6

Either n=4x+a OR n=4x+a, where x is positive integer.... AND a= 1 or 3
Now,
2^(4x+a) + 3^(4x+a) = y*125, where y is an odd number

Answer 7

Am I on the right track? @mukushla

Answer 8

sorry i lost connection...this one needs a little number theory... :/ im going to close this and i'll post some other one's so that u just need to apply elementary methods :)

Answer 9

@mukushla CAN U PLZ GIVE THE SOLUTION

Answer 10

I would really like to see this

Answer 11

yeah i will :) plz wait a min

Answer 12

ok

Answer 13

forgot to say this needs modular arithmetic but i will post solution if u want :)

Answer 14

if n is odd 2^n+3^n is divisible by 5 so \[2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1}\]must be divisible by 25 but firstly it must be divisible by 5

Answer 15

in simple words when we divide 2 by 5 remainder is 2 and when we divide -3 by 5 remainder is 2 again so\[2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1}\equiv n\times2^n \ \ \text{mod} \ 5\]

Answer 16

actually here n2^n is remainder when we divide \[2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1}\]by 5 ( i changed -3 to 2)

Answer 17

it will be very helpful for u try to work it out

Answer 18

so n must be a multiple of 5 so let n=5k

Answer 19

a similar proccess will show that k=5m and smallest will be for m=1 and n=25

Answer 20

Thanx