fun question :)

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fun question :)

Mathematics
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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.

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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

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Other answers:

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 ?
good :)
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
yes :D
(1,2,4,3) is the last one
yes ᗡ:
should i post the solution? or some hints? :)
no wait :)
ok (:
\(p!+q!+r!=3^s\) 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).

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