A community for students.
Here's the question you clicked on:
 0 viewing
mukushla
 3 years ago
Find the smallest positive integer for which\[5^32^n+3^n\ .\]
mukushla
 3 years ago
Find the smallest positive integer for which\[5^32^n+3^n\ .\]

This Question is Closed

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2\(n\) must be odd so \[2^n+3^n=5(2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1})\]

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0Can u PLZ explain the question

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0Either 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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0Am I on the right track? @mukushla

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2sorry 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 :)

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla CAN U PLZ GIVE THE SOLUTION

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0I would really like to see this

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2yeah i will :) plz wait a min

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2forgot to say this needs modular arithmetic but i will post solution if u want :)

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2if n is odd 2^n+3^n is divisible by 5 so \[2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1}\]must be divisible by 25 but firstly it must be divisible by 5

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2in simple words when we divide 2 by 5 remainder is 2 and when we divide 3 by 5 remainder is 2 again so\[2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1}\equiv n\times2^n \ \ \text{mod} \ 5\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2actually here n2^n is remainder when we divide \[2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1}\]by 5 ( i changed 3 to 2)

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2it will be very helpful for u try to work it out

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2so n must be a multiple of 5 so let n=5k

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2a similar proccess will show that k=5m and smallest will be for m=1 and n=25
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.