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## anonymous 4 years ago Find the smallest positive integer for which$5^3|2^n+3^n\ .$

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1. anonymous

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

2. anonymous

Can u PLZ explain the question

3. anonymous

@mukushla

4. anonymous

@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

5. anonymous

thanx @mukushla

6. anonymous

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

7. anonymous

Am I on the right track? @mukushla

8. anonymous

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 :)

9. anonymous

@mukushla CAN U PLZ GIVE THE SOLUTION

10. anonymous

I would really like to see this

11. anonymous

yeah i will :) plz wait a min

12. anonymous

ok

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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)

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

Thanx

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