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

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\(n\) must be odd so \[2^n+3^n=5(2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1})\]
Can u PLZ explain the question

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Other answers:

@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
thanx @mukushla
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
Am I on the right track? @mukushla
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 :)
@mukushla CAN U PLZ GIVE THE SOLUTION
I would really like to see this
yeah i will :) plz wait a min
ok
forgot to say this needs modular arithmetic but i will post solution if u want :)
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
in simple words when we divide 2 by 5 remainder is 2 and when we divide -3 by 5 remainder is 2 again so\[2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1}\equiv n\times2^n \ \ \text{mod} \ 5\]
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)
it will be very helpful for u try to work it out
so n must be a multiple of 5 so let n=5k
a similar proccess will show that k=5m and smallest will be for m=1 and n=25
Thanx

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