Every positive integer with exactly 91 divisors is a perfect 6-th power???

Discussion in 'General Math' started by don.lotto, Sep 6, 2006.

1. don.lottoGuest

Every positive integer with exactly 91 divisors is a perfect 6-th
power???

6th root(7/9 *40!( 6!*34!) ) = ???

don.mcdonald nz 6/sept 2006.

don.lotto, Sep 6, 2006

2. don.lottoGuest

wrote:

google calc doesnot work any more xx

factorise etc.

don.mcdonald.

don.lotto, Sep 6, 2006

3. Ed MurphyGuest

Yes, because it must be either (p1^6 * p2^12) or (p1^90).
2985406 and two-thirds.

2985406 has four divisors (1, 2, 149203, 2985406).

2985407 has four divisors (1, 107, 27901, 2985407).

Why do you expect 7/9 * C(40,6) to be related to a number with 91
divisors? I don't get it.

Ed Murphy, Sep 6, 2006
4. Ed MurphyGuest

Google Calculator handles factorials just fine:

(7 / 9) * ((40 !) / ((6 !) * (34 !))) = 2 985 406.67

((7 / 9) * ((40 !) / ((6 !) * (34 !))))^(1 / 6) = 11.9996133

Perhaps you botched the parentheses? I did, the first time.

This is indeed close to 2^12 * 3^6 = 2985984 (a difference of 577 1/3).

Ed Murphy, Sep 6, 2006
5. PubkeybreakerGuest

Yes. 91 = 7*13, so any number with 91 divisors is of the
form p^6 q^12 or r^90.

Pubkeybreaker, Sep 6, 2006
6. don.lottoGuest

well, 6th root didnot kick in the calc.

sometimes 3*5 does not kick in the calc.
something like that.

ouch 40! / (...) xxxxxxxxx division may be missing.

((7 / 9) * (40 choose 6))^(1 / 6) = 11.9996133

thanks Ed. and pub key.

It seemed to be much freer with respect to parentheses in the past.

i will keep trying. but it haS BEEN harder and less useable lately.
don.mcdonald.

don.lotto, Sep 7, 2006
7. don.lottoGuest

....
Re: Every positive integer with exactly 91 divisors is a perfect 6-th
power???

((7 / 9) + (1 / 6 648)) * (40 choose 6) = 2 985 984.04

a 7-digit phone number. (04) = wellington and kapiti new zealand.

= 12^ 6th.
the smallest positive integer with 91 divisors.

I am struggling with calc google.

greater than.
An estimate of its magnitude. cheers,

don.mcdonald.
author's solution soon.

don.lotto, Sep 7, 2006
8. bill.dalyGuest

Exactly, but it is possible to improve on this. If d is a product of
primes, each of which is of the form 6n+1, then any number with exactly
d divisors is a 6th power. For that matter, if m is any number, and d
is a product of primes, each of which leaves a residue of 1 mod m, then
any number with exactly d divisors is an mth power.

bill.daly, Sep 13, 2006