[math-fun] 2015 with the other six digits
Hello Math-funsters, what could be an economic way to get 2015 with the six other digits [3,4,6,7,8,9] and the usual signs [=,+,-,x,:,^,...]? Best, É. Catapulté de mon aPhone
Assuming you mean using the numbers once only: (6^3 + 7)*9 + 8 or (8^3 - 6)*4 - 9 On 21 Dec 2014, at 11:15, Eric Angelini wrote:
Hello Math-funsters, what could be an economic way to get 2015 with the six other digits [3,4,6,7,8,9] and the usual signs [=,+,-,x,:,^,...]? Best, É. Catapulté de mon aPhone _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The meaning and purpose of life is to give life purpose and meaning. The instigation of violence indicates a lack of spirituality.
Many thanks to all -- and Fred and Erich Friedman (private mail)
(69 - 4) x (38 - 7) = 2015
Best, É. Catapulté de mon aPhone
Le 21 déc. 2014 à 15:36, "Fred W. Helenius" <fredh@ix.netcom.com> a écrit :
On 12/21/2014 6:15 AM, Eric Angelini wrote:
Hello Math-funsters, what could be an economic way to get 2015 with the six other digits [3,4,6,7,8,9] and the usual signs [=,+,-,x,:,^,...]?
(69-4)*(38-7)
-- Fred W. Helenius fredh@ix.netcom.com
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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! ??? --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
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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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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (8)
-
Daniel Asimov -
David Makin -
Eric Angelini -
Fred W. Helenius -
Hans Havermann -
Michael Greenwald -
Olivier Gerard -
Simon Plouffe