Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mukushla Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \(n\) 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 years ago
  2. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Can u PLZ explain the question

    • 2 years ago
  3. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @mukushla

    • 2 years ago
  4. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @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

    • 2 years ago
  5. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thanx @mukushla

    • 2 years ago
  6. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 2 years ago
  7. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Am I on the right track? @mukushla

    • 2 years ago
  8. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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 :)

    • 2 years ago
  9. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @mukushla CAN U PLZ GIVE THE SOLUTION

    • 2 years ago
  10. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I would really like to see this

    • 2 years ago
  11. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah i will :) plz wait a min

    • 2 years ago
  12. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

    • 2 years ago
  13. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  14. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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

    • 2 years ago
  15. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  16. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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)

    • 2 years ago
  17. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  18. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  19. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  20. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanx

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.