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imqwerty
 one year ago
fun question :)
imqwerty
 one year ago
fun question :)

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2Find all 4 pairs (p,q,r,s) of natural numbers such that \[p \le q \le r\] and \[p!+q!+r!=3^s\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dude don't give away hints 2 minutes after posting the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this isn't a very fun question

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5this is indeed a fun question, the hints only act as spoilers for those who genuinely want to try

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i didnt saw any hint ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5It seems (1,1,1,1) is the only solution... I kinda have a proof too but I'll let others try :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2there are 3 solutions :) and this is one of them

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5right, (1,2,3,2) is the second one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5(1,2,4,3) is the last one

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2should i post the solution? or some hints? :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(p!+q!+r!=3^s\) 1) Clearly \(p=1\) (why ?) 2) so we look at \(q!+r!=3^s1\) 3) since the right hand side is not divisible by \(3\), it must be the case that \(q=2\) for all \(r\gt 1\). 4) now we look at \(r! = 3^s3\) 5) Clearly this equation has no solutions for \(r\ge 9\) and \(s\ge 2\) because the right hand side wont be divisible by \(9\).

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes correct :) our solutions are the same

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5please do post the solution for the completeness sake..

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2Note that if p > 1 then the lefthand side is even, and therefore p = 1. If q > 2 then 3 divides q!+r! and hence 3 does not divide the lefthand side. Therefore q = 1 or q = 2. If q = 1 then r! + 2 = 3^s, so r < 2 and hence s = 1. If q = 2 then r! = 3^s 3. Note that s = 1 does not give any solution. If s> 1 then 9 does not divide r!, so r < 6. By checking the values for r = 2, 3, 4, 5 we see that r = 3 and r = 4 are the only two solutions. Thus (p, q, r, s) = (1, 1, 1, 1), (1, 2, 3, 2) or (1, 2, 4, 3).
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