anonymous
  • anonymous
Find the smallest positive integer for which\[5^3|2^n+3^n\ .\]
Meta-math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\(n\) must be odd so \[2^n+3^n=5(2^{n-1}-3\times 2^{n-2}+3^2 \times 2^{n-3}+... -2\times3^{n-2}+3^{n-1})\]
anonymous
  • anonymous
Can u PLZ explain the question
anonymous
  • anonymous
@mukushla

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.