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mukushla

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

  • one year ago
  • one year ago

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

    • one year ago
  2. sauravshakya
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    Can u PLZ explain the question

    • one year ago
  3. sauravshakya
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    @mukushla

    • one year ago
  4. mukushla
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    @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

    • one year ago
  5. sauravshakya
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    thanx @mukushla

    • one year ago
  6. sauravshakya
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    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

    • one year ago
  7. sauravshakya
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    Am I on the right track? @mukushla

    • one year ago
  8. mukushla
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    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 :)

    • one year ago
  9. sauravshakya
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    @mukushla CAN U PLZ GIVE THE SOLUTION

    • one year ago
  10. sauravshakya
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    I would really like to see this

    • one year ago
  11. mukushla
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    yeah i will :) plz wait a min

    • one year ago
  12. sauravshakya
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    ok

    • one year ago
  13. mukushla
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    forgot to say this needs modular arithmetic but i will post solution if u want :)

    • one year ago
  14. mukushla
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    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

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

    • one year ago
  16. mukushla
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    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)

    • one year ago
  17. mukushla
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    it will be very helpful for u try to work it out

    • one year ago
  18. mukushla
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    so n must be a multiple of 5 so let n=5k

    • one year ago
  19. mukushla
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    a similar proccess will show that k=5m and smallest will be for m=1 and n=25

    • one year ago
  20. sauravshakya
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    Thanx

    • one year ago
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