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by grockit_jake » Tue Sep 29, 2009 11:57 am
Yes, you are correct. You can first break 130 into its factors, and then recombine to see how many fit.

130 = 2*65 = 2*5*13

2, 5, 13, 10, 65, 26, 130, 1.

Count your way up, and when you hit the max, you know you are done.
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by punitkaur » Tue Sep 29, 2009 12:00 pm
thanks for confirming. but the way i did it was a little different.
130
prime factors - 2,5,13.

no of divisors is multiplication of (powers of each +1)
2x2x2 = 8.

If we got a larger number then this method is easier I believe?
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by tohellandback » Tue Sep 29, 2009 8:11 pm
punitkaur wrote:thanks for confirming. but the way i did it was a little different.
130
prime factors - 2,5,13.

no of divisors is multiplication of (powers of each +1)
2x2x2 = 8.

If we got a larger number then this method is easier I believe?
I think your method is better than one stated above..
the other method can be time-taking if there are large numbers.
The powers of two are bloody impolite!!
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Thanks!

by iships » Wed Sep 30, 2009 12:59 am
I usually get stuck in such "How many" questions,,,
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Can you explain that strategy above?

by spartigrl » Sat Oct 03, 2009 10:46 am
How can you determine from the prime factors how many total factors there are?
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