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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 00:36
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A number is a product of 5 prime factors, 2 of them are same. How many factors does it have?



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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 01:13
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n = a^2bcd
Number of factors = (3)(2)(2)(2) = 24?
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 01:34
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Hmmm, I get 18 factors

I used the prime numbers
2 * 2 * 3 * 5 * 7 = 420

I try to made all possible combinations by combining all possible prime numbers.

Please correct me if I am wrong.

Regards,

Alex
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 01:37
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A number is a product of 5 prime factors, 2 of them are same.
For explanation, I considered these to be a, b, c, d. n = a^2bcd a is reapeted twice.

To find the number of factors, say for x^a.y^b.z^c, simply multiply the powers+1, i.e. no of factors = (a+1)(b+1)(c+1)

In this case, no of factors of a^2bcd = 3.2.2.2 = 24

I learnt this formula from the forum - an old discusion between Stolyar and AkamaiBrah. That said, beware of its use, it can be used only in the simple case.
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 04:49
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i writed all of them and i obtained 23 factors
+ 1 which is always a factor

1 a b c d a^2 ab ac ad bc bd cd a^2b a^2c a^2d acb adc adb bcd a^2cb a^2dc a^2db abcd a^2bcd

24

Sounds like this formula is working very well... I will use it from now on !
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 04:59
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I did an recount and came also to 24.

I dont know why I came to 18 in the first place!!!

However is it also possible to use some combination theory
2 2 3 5 7 = 5 prime numbers.
2 are the same which can be counted as 1

Therefore 4*3*2*1 = 24 possibilities.

It is correct to use this method???


Regards,

Alex
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 08:04
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hardworker_indian
A number is a product of 5 prime factors, 2 of them are same.
For explanation, I considered these to be a, b, c, d. n = a^2bcd a is reapeted twice.

To find the number of factors, say for x^a.y^b.z^c, simply multiply the powers+1, i.e. no of factors = (a+1)(b+1)(c+1)

In this case, no of factors of a^2bcd = 3.2.2.2 = 24

I learnt this formula from the forum - an old discusion between Stolyar and AkamaiBrah. That said, beware of its use, it can be used only in the simple case.
Show more

Yes, that discussion was gold. I learned it from this forum too.
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A number is a product of 5 prime factors, 2 of them are 17 Sep 2004, 10:18
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Thanks hardworker_indian for the formula and explanation.



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