Positive integers( A short question )

[primarygun]

How many positive integers less than 500 have exactly 15 positive integer factors?
I know the answer, but not sure it. Can you give me the answer ?

[Tide]

HINT: 2^{15} = 32768

[matt grime]

I'm not sure I get the hint. 2.3.5.7 has 15 positive integer factors (excluding itself).

[Tide]

I'm not sure I get the hint. 2.3.5.7 has 15 positive integer factors (excluding itself).

Oh, I took "has 15 postive factors" as referring to prime factorization. (2x3x5x7 has 4 prime factors!)

Apparently that's not what primarygun was looking for.

[matt grime]

I'm not sure if we're supposed to include 1 and itself to be honest. If we are then the only numbers are those... actually, I'm sorry, I always get suspicious when people say "i've got the answer" but don't say what it is. Could you explain how you got your answer, Primarygun, as i don't want to give it away too easily?

[shmoe]

It's more natural to include the number itself and 1 when counting its number of positive divisors (it makes for a multiplicative function that way), and this is how I would interpret the question. Clarification wouldn't hurt of course.

Some simple hints in any case. If a number has 15 divisors, what does this say about it's prime factorization? Specifically, can you say how many distinct prime factors it has? What can you say about the exponents appearing in the prime factorization?

[CTS]

Doesn't

2 * 3 * 5 * 7

have

\tau (2 * 3 * 5 * 7) = \tau (2) * \tau (3) * \tau (5) * \tau (7) = 2 * 2 * 2 * 2 = 16

positive factors?

I think what you need to look at is the factors of 15: 1, 3, 5, and 15

15 = 1 * 15


15 = 3 * 5

The first case gives

\tau (p^{1-1} * q^{15-1}) = \tau (q^{14})

Because

2^{14} > 500

Move on to the second case

\tau (p^{3-1} * q^{5-1}) = \tau (q^4 * p^2) = 15

Move through the primes until you surpass 500:

2^4 * 3^2 = 144 < 500


\tau (144) = 15

[shmoe]

So it looks like 144 is the only number less than 500 with 15 positive factors, correct?

No, it's not the only one. Try other values of p, q. Let's not give too much away though?


ps.when Matt said 2.3.5.7 had 15 positive divisors, note he also said "excluding iself".

[primarygun]

Sorry, haven't been online so long.
I guess it should be 3.
As you see, 15 only can be expressed as 1x15 or 3x5 for positive interger factor.
Let the number have 15 factors be n=pq (p and q : positive integers)
For 15 factor, p=1, q^14=n (q : prime number) or p^2q^4=n (p , q : prime number)
But 2^14> 500, so the first expression is rejected.
p^2 q^4 = n
Find the limit for both first, start to do it with q.
So q=2, q^4=16 n=500
So the possible largest integer is 5.
Select q from 2,3 Select p from 2,3,5.
So answer: 400, 144, and 162

[shmoe]

So answer: 400, 144, and 162

Method looks good, but check how you ended up with 162.

[dextercioby]

Method looks good, but check how you ended up with 162.

324 has 17 positive divisors.Excluding 1 and 324,it's left with 15.

Daniel.

[shmoe]

324 has 17 positive divisors.Excluding 1 and 324,it's left with 15.

Daniel.

Nope. The formula for the number of divisors that's been kicking around the last few posts for a product of two primes, \tau(p^n q^m)=(n+1)(m+1) includes 1 and itself in the count.