How many numbers under 2010 have only 3 factors?

I got this question recently, I completed it but I do not know if it is correct. Could someone work it out?

Thanks!

Printable View

- Aug 5th 2010, 01:42 AMjgv115How many numbers under 2010....
How many numbers under 2010 have only 3 factors?

I got this question recently, I completed it but I do not know if it is correct. Could someone work it out?

Thanks! - Aug 5th 2010, 01:47 AMBacterius
That would be 503 according to me. Regardless of whether the factors are distinct or not. Sorry, checked by computer, no real maths here.

- Aug 5th 2010, 04:32 AMjgv115
mm.. How did you get that? I only got 14.

I figured out that only the square of prime numbers have 3 factors.

eg. $\displaystyle 5^2 = 25 $ 25 has 1,5 and 25

and so on... - Aug 5th 2010, 04:47 AMBacterius
$\displaystyle 3 \times 5 \times 7 = 105$, this number has three factors ... I think you have mixed up factors and divisors.

- Aug 5th 2010, 06:20 AMPlato
- Aug 5th 2010, 06:26 AMundefined
Supposing there's room for interpretation, it could be one of these three, listed alphabetically

1) How many distinct prime factors in the prime factorisation?

2) How many divisors?

3) How many not-necessarily-distinct prime factors in the prime factorisation?

Personally I would like the question to be more specific, but others may claim it is clearly one of these three and not the other two.

Edit: and a fourth

4) How many proper divisors?

(and apparently "d" comes before "n".... chalk it up to being groggy in the morning) - Aug 5th 2010, 06:34 AMPlato
I think the term

*factor*well defined.

A common counting question is "How many factors does 1224 have?" - Aug 5th 2010, 06:41 AMundefined
You're right, I'm just more used to seeing the word divisor used. I also interpreted the question as asking how many divisors upon my first reading but didn't want to discredit Bacterius's answer. But if the question wants prime factors, it should say prime factors.

- Aug 5th 2010, 06:51 AMAlso sprach Zarathustra
(304 choose 3)-something=:)

- Aug 5th 2010, 06:59 AMundefined
- Aug 5th 2010, 11:54 AMBacteriusQuote:

You're right, I'm just more used to seeing the word divisor used. I also interpreted the question as asking how many divisors upon my first reading but didn't want to discredit Bacterius's answer. But if the question wants prime factors, it should say prime factors.

Sorry - Aug 5th 2010, 12:23 PMPlato
I don’t think that understanding of

*factor*is historically accurate.

Here is a good statement on that topic. - Aug 5th 2010, 12:28 PMBacterius
Going by the same token, I don't think that understanding of factor is "modernly" accurate, based on my reading over the past few months. But I don't want to fight over words, we're on a maths forum and I guess either the OP wasn't clear enough, either I was just too stupid to hit the point.

- Aug 5th 2010, 09:34 PMUsername
This exact question was in the Australian Maths Competition, the answer is 14.

- Aug 5th 2010, 10:26 PMchisigma
If we consider 'three factors' as 'three distinct prime factors' [so that, for example, 1 is not prime number and is not a 'factor'...] a simple systematic search can be done...

$\displaystyle 2 \times 3 \times 5 = 30$

$\displaystyle 2 \times 3 \times 7 = 42$

$\displaystyle 2 \times 5 \times 7 = 70$

$\displaystyle 2 \times 3 \times 11 = 66$

$\displaystyle 2 \times 5 \times 11 = 110$

$\displaystyle 2 \times 7 \times 11 = 154$

$\displaystyle 2 \times 3 \times 13 = 78$

$\displaystyle 2 \times 5 \times 13 = 130$

$\displaystyle 2 \times 7 \times 13 = 182$

$\displaystyle 2 \times 11 \times 13 = 286$

$\displaystyle 2 \times 3 \times 17 = 102$

$\displaystyle 2 \times 5 \times 17 = 170$

$\displaystyle 2 \times 7 \times 17 = 238$

$\displaystyle 2 \times 11 \times 17 = 374$

... so that 14 different numers less than 2010 expressed as product of three distinct prime fators have been found till now. But we can proceed futher...

$\displaystyle 2 \times 13 \times 17 = 442$

$\displaystyle 2 \times 3 \times 19 = 114$

... and so on...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$