Find all 4 pairs (p,q,r,s) of natural numbers such that \[p \le q \le r\] and \[p!+q!+r!=3^s\]
dude don't give away hints 2 minutes after posting the question
lol ok cx
this isn't a very fun question
this is indeed a fun question, the hints only act as spoilers for those who genuinely want to try
i didnt saw any hint ?
It seems (1,1,1,1) is the only solution... I kinda have a proof too but I'll let others try :)
there are 3 solutions :) and this is one of them
right, (1,2,3,2) is the second one
(1,2,4,3) is the last one
should i post the solution? or some hints? :)
no wait :)
\(p!+q!+r!=3^s\) 1) Clearly \(p=1\) (why ?) 2) so we look at \(q!+r!=3^s-1\) 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 \(r! = 3^s-3\) 5) Clearly this equation has no solutions for \(r\ge 9\) and \(s\ge 2\) because the right hand side wont be divisible by \(9\).
yes correct :) our solutions are the same
please do post the solution for the completeness sake..
ok wait a min :)
Note that if p > 1 then the left-hand side is even, and therefore p = 1. If q > 2 then 3 divides q!+r! and hence 3 does not divide the left-hand 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).