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anonymous
 4 years ago
Find the smallest positive integer for which\[5^32^n+3^n\ .\]
anonymous
 4 years ago
Find the smallest positive integer for which\[5^32^n+3^n\ .\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(n\) must be odd so \[2^n+3^n=5(2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1})\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i will :) plz wait a min

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forgot to say this needs modular arithmetic but i will post solution if u want :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in simple words when we divide 2 by 5 remainder is 2 and when we divide 3 by 5 remainder is 2 again so\[2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1}\equiv n\times2^n \ \ \text{mod} \ 5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it will be very helpful for u try to work it out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so n must be a multiple of 5 so let n=5k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a similar proccess will show that k=5m and smallest will be for m=1 and n=25
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