here is another one. 3737 = 2015 if 3737 is in base 8. and of course merry 31 oct, which is 25 DEC. Best regards, Simon Plouffe 2014-12-24 13:04 GMT+01:00 Olivier Gerard <olivier.gerard@gmail.com>:
But you have another, related, interesting case :
23 ! = 5 * 6 * ... * 24
and in general
(n!-1)! = (n!)! / n!
and if you generalize to
q*(q+k)*(q+2*k)...(q+n*k) = r*(r+d)*(r+2*d)*(r+3*d)...(r+m*d)
you have many instances of product coincidences.
On Wed, Dec 24, 2014 at 5:55 AM, Michael Greenwald < greenwald@cis.upenn.edu> wrote:
On 2014-12-23 20:27, Daniel Asimov wrote:
Nice one!
I hadn't noticed before that 7*8*9*10 = 7!.
Are there any other numbers q for which there is a b > 0 such that
q*(q+1)*...*(q+b) = q!
It seems unlikely: q+1 through q+b must all be composite (else could not = (q-1)! ) But they also must contain multiples of every prime < q. If p is the largest prime < q, then there can be no primes between p and 2p (other than possibly q). Since 11 is the 2nd Ramanujan Prime, all n > 11 has >= 2 primes between n and 2n. But 11 can't be in [q+1, q+b] because
it
is prime. So 7 is the largest q that this holds for.
???
--Dan
On Dec 21, 2014, at 2:17 PM, Hans Havermann <gladhobo@teksavvy.com>
wrote:
Don't forget to count down on New Year's Eve:
10*9*8*7/6/5*4*3-2+1
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