Project Euler Forum

Where can i find that PDF

akshayms wrote:Where can i find that PDF

After you solve the problem, there will be a PDF icon besides the forum icon. This links to the PDF file.

my program(in java) is too slow to find the first factor with over 500 divisors. right now i have two functions, one that finds the triangle number, and one that finds every factor of a number, then returns how many were found.

my triangle function is nearly instant, since it just uses an equation. my factoring function, tests every number less than or equal to the number being factored, and counts how many were perfectly divisible.

does anyone know a quicker way to find how many factors a number has? the current way ios far too slow for large numbers and i cant find the answer in under a minute, as i should be able to.

Please don't start a new topic for a problems if such a topic already exists.

m34tcode wrote:my program(in java) is too slow to find the first factor with over 500 divisors.

... Other comments removed ...

does anyone know a quicker way to find how many factors a number has?

A factor is not a divisor. I would suggest reviewing your program to ensure that you are solving the requested problem. A brute force program can solve the problem in a very short time.

My solution is in Java and completes in .031 sec.

davidFashion wrote:A factor is not a divisor.

Factor and divisor are synonyms and mean exactly the same thing.

To m34tcode - I'm afraid this forum isn't for giving away hints; you'll need to come up with the idea for something faster yourself.

TripleM wrote:

davidFashion wrote:A factor is not a divisor.

Factor and divisor are synonyms and mean exactly the same thing.

Yes, you are correct. When I wrote "factor", I was thinking "prime factor". Thanks for the clarification.

m34tcode wrote:my triangle function is nearly instant, since it just uses an equation. my factoring function, tests every number less than or equal to the number being factored, and counts how many were perfectly divisible.

does anyone know a quicker way to find how many factors a number has?

Yes, there is a faster way, and most of the successful problem 12 solvers probably used it (I know I did).

Your challenge is to find it.

Hi, I tried the problem 12,
i got one answer, which seems to be incorrect according to the website.
the number has 24 and 23 divisors each. can anyone give any further hint.

akv1kor wrote:Hi, I tried the problem 12,
i got one answer, which seems to be incorrect according to the website.
the number has 24 and 23 divisors each. can anyone give any further hint.

4 has 3 divisors.
6 has 4 divisors.

But...
4*6 = 24 does not have 12 divisors, only 8.

thundre wrote: 4 has 3 divisors.
6 has 4 divisors.

But...
4*6 = 24 does not have 12 divisors, only 8.

Here we have N&N+1 so above logic does not fit i suppose.
apart from 1,other divisors are mutually exclusive.
my number has 577 divisors,Is this any way near.

Unless I'm mistaken the smallest number having 577 divisors is 2⁵⁷⁶ (577 is prime).
However, if you multiply two coprime numbers with 23 and 24 divisors the number of divisors would be 23*24=552.
The number you are looking for has 576 divisors.
Do you handle the division by two correctly?

its my mistake actually i did a mistake in adding 1 as a divisor for the first number.
now the number has 24 and 24 divisors each and i give (n*(n+1))/2 as solution but still i have wrong answer.
***0*1*0 is my value for ****th triangle number. pls match asterik with answers and advice

I'm not sure what your problem is, but let me give two examples:

suppose n=3, then n+1=4, so n(n+1)/2=6.
6 has 4 divisors: 1,2,3, and 6.
How can I calculate this from 3 and 4?
The numers to multiply are 3 and 4/2=2.
3 has two divisors: 1 and 3, and 2 has two divisors: 1 and 2.
So the number if divisors of 6 is 2*2=4.

If we take n=8 then n+1=9
So we have to multiply 8/2=4 and 9, makes 36.
36 has 9 divisors 1,2,3,4,6,9,12,18 and 36.
4 and 9 have both 3 divisors and 3*3=9.

2 and 3 are coprime.
4 and 9 are coprime.
It's important !

Of course, but akv1kor knows that already.

Sorry, it was written just before.

thanks HK for your kind advice, you got it correct, i never thought to find factors of n/2
now i got it

akv1kor wrote:thanks HK for your kind advice, you got it correct, i never thought to find factors of n/2
now i got it

You're welcome.

Oh man, what a horrible time to forget to write in a pause function at the end of the program!!!!