# Thread: How many numbers under 2010....

1. ## How 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!

2. That would be 503 according to me. Regardless of whether the factors are distinct or not. Sorry, checked by computer, no real maths here.

3. mm.. How did you get that? I only got 14.

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

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

and so on...

4. $3 \times 5 \times 7 = 105$, this number has three factors ... I think you have mixed up factors and divisors.

5. Originally Posted by Bacterius
$3 \times 5 \times 7 = 105$, this number has three factors ... I think you have mixed up factors and divisors.
Actually $105$ has $2\cdot 2 \cdot 2=8$ factors: $1,~3,~5,~7,~15,~21,~35,~\&~105.$.

Originally Posted by jgv115
mm..
I figured out that only the square of prime numbers have 3 factors.

eg. $5^2 = 25$ 25 has 1,5 and 25
That is correct. Yes 14

6. Originally Posted by jgv115
How many numbers under 2010 have only 3 factors?
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)

7. I think the term factor well defined.
A common counting question is "How many factors does 1224 have?"

8. Originally Posted by Plato
I think the term factor well defined.
A common counting question is "How many factors does 1224 have?"
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.

9. (304 choose 3)-something=

10. Originally Posted by Also sprach Zarathustra
(304 choose 3)-something=

11. 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.
I agree with Undefined, "factor" is just too close to "prime factor". And it seemed more interesting
Sorry

12. Originally Posted by Bacterius
I agree with Undefined, "factor" is just too close to "prime factor". And it seemed more interesting.
I don’t think that understanding of factor is historically accurate.
Here is a good statement on that topic.

13. 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.

14. This exact question was in the Australian Maths Competition, the answer is 14.

15. 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...

$2 \times 3 \times 5 = 30$

$2 \times 3 \times 7 = 42$

$2 \times 5 \times 7 = 70$

$2 \times 3 \times 11 = 66$

$2 \times 5 \times 11 = 110$

$2 \times 7 \times 11 = 154$

$2 \times 3 \times 13 = 78$

$2 \times 5 \times 13 = 130$

$2 \times 7 \times 13 = 182$

$2 \times 11 \times 13 = 286$

$2 \times 3 \times 17 = 102$

$2 \times 5 \times 17 = 170$

$2 \times 7 \times 17 = 238$

$2 \times 11 \times 17 = 374$

... 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...

$2 \times 13 \times 17 = 442$

$2 \times 3 \times 19 = 114$

... and so on...

Kind regards

$\chi$ $\sigma$

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