Hi everybody,
What is the smallest positive integer with exactly 768 divisors?
I think that 2^767 might have 768 divisors, but I think there would be a smaller one.
I'm thinking that it would be some product of prime numbers.
Like etc
Anyone got any ideas.
Thanks.
Hi Romsek,
it is really nice to 'see' you again.
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I do mean unique.
Think about 2^5. By my logic this should have 6 distict divisors
1*2*2*2*2*2
1,2,4,8,16,32 See it has 6 distict divisors.
The reason I know that this is not the smallest one is demonstrated as follows
2^3=8, this has 4 factors, 1,2,4,and 8
but
6 is a smaller number with 4 factors, that is 1,2,3, and 6
If you were to use the suggested a divisor would be the product of 0,1,2,... or x 2's, 0,1,2,... or y 3's and 0,1,2,... or z 5's. That would give you a possible (x+1)(y+1)(z+1) divisors. You would need that to equal 768.
I think though it's necessary to include higher primes.
If we assume that the divisors are of the form then we would need
Since the best that I can come up with for the moment, subject to further investigation, is,
which produces the number
Hi BobP,
I wasn't suggesting that 5 would be the highest prime, (I actually put, etc, in my question)
I've got to think about your answer. I don't understand the logic you have used. I'm tired now. It might make more sense tomorrow.
Does your list really have 768 distinct combinations. (I hope i got my wording right that time) I did a quick calculation of my own and it could do. Still, I thought there would be more low prime numbers than that.
I'll be back
Thanks
Melody.
Building from BobP's good work, I think a better mix of values for a, b, c, etc are:
a=3, b=3, c= 2, d=1, e=1, f=1, and g=1. This gives:
with total number of factors = 4 x 4 x 3 x 2 x 2 x 2 x 2 = 768.
The strategy is to use higher powers for lower primes, as long as the prime raised to that value is less than the next highest prime that would otherwise be needed. For example 2^3 = 8 is smaller than adding more primes (19, 23, etc). It would be nice to be ableto use 2^4, but then (a+1) = (4+1) = 5, and that does not divide into 768, so we/re limited to 2^3. Notice that one of the divisors of 768 is 3, so one of the primes is going to have to have a power of 2. So we need to see which works better: 3^2 and include 19 or 3^3, 5^2 and not include 19. I foind the latter gives a lower value.