# What is the smallest positive integer with exactly 768 divisors?

• Jan 21st 2014, 10:57 PM
Melody2
What is the smallest positive integer with exactly 768 divisors?
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 $\displaystyle \:2^x\times3^y\times5^z\;$ etc
Anyone got any ideas.

Thanks.

• Jan 21st 2014, 11:17 PM
romsek
Re: What is the smallest positive integer with exactly 768 divisors?
Quote:

Originally Posted by Melody2
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 $\displaystyle \:2^x\times3^y\times5^z\;$ etc
Anyone got any ideas.

Thanks.

I would think it's the product of the first 768 primes... 2*3*5*7*11*13.... *5489

You do mean unique divisors I assume. Otherwise i guess it would be 2^768.
• Jan 21st 2014, 11:29 PM
Melody2
Re: What is the smallest positive integer with exactly 768 divisors?
Hi Romsek,
it is really nice to 'see' you again.
-------------------------------------------------------------------
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
• Jan 22nd 2014, 03:31 AM
BobP
Re: What is the smallest positive integer with exactly 768 divisors?
If you were to use the suggested $\displaystyle 2^{x}.3^{y}.5^{z},$ 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 $\displaystyle 2^{a}.3^{b}.5^{c}.7^{d}.\dots$ then we would need

$\displaystyle (a+1)(b+1)(c+1)(d+1) \dots = 768.$

Since $\displaystyle 768=2^{8}.3,$ the best that I can come up with for the moment, subject to further investigation, is,

$\displaystyle a=2, b=c=d=e=f=g=h=I=1$ which produces the number $\displaystyle 2^{2}.3.5.7.11.13.17.19.23=446185740.$
• Jan 22nd 2014, 04:17 AM
Melody2
Re: What is the smallest positive integer with exactly 768 divisors?
Quote:

Originally Posted by BobP
If you were to use the suggested $\displaystyle 2^{x}.3^{y}.5^{z},$ 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 $\displaystyle 2^{a}.3^{b}.5^{c}.7^{d}.\dots$ then we would need

$\displaystyle (a+1)(b+1)(c+1)(d+1) \dots = 768.$ WHY?

Since $\displaystyle 768=2^{8}.3,$ the best that I can come up with for the moment, subject to further investigation, is,

$\displaystyle a=2, b=c=d=e=f=g=h=I=1$ which produces the number $\displaystyle 2^{2}.3.5.7.11.13.17.19.23=446185740.$

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 (Sleepy)
Thanks
Melody.
• Jan 22nd 2014, 05:42 AM
ebaines
Re: What is the smallest positive integer with exactly 768 divisors?
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:

$\displaystyle 2^3 \times 3^3 \times 5^2 \times 7 \times 11 \times 13 \times 17 = 91891800.$

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.
• Jan 22nd 2014, 08:26 AM
BobP
Re: What is the smallest positive integer with exactly 768 divisors?
$\displaystyle 2^{5}\times 3^{3}\times 5 \times 7 \times 11 \times 13 \times 17 = 73513440,$

has

$\displaystyle 6 \times 4 \times 2 \times 2 \times 2 \times 2 \times 2 = 768$

divisors.

Any further offers ?
• Jan 26th 2014, 03:12 AM
Melody2
Re: What is the smallest positive integer with exactly 768 divisors?
I believe it all makes sense. Thanks everyone. (Rofl)