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mukushla
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Find the smallest positive integer for which\[5^32^n+3^n\ .\]
 2 years ago
 2 years ago
mukushla Group Title
Find the smallest positive integer for which\[5^32^n+3^n\ .\]
 2 years ago
 2 years ago

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Can u PLZ explain the question
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
@mukushla
 2 years ago

mukushla Group TitleBest 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
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
thanx @mukushla
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.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

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
if 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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.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\[2^{n1}3\times 2^{n2}+3^2 \times 2^{n3}+... 2\times3^{n2}+3^{n1}\equiv n\times2^n \ \ \text{mod} \ 5\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
actually 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)
 2 years ago

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
a similar proccess will show that k=5m and smallest will be for m=1 and n=25
 2 years ago
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