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Divisors

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gmatt14
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Divisors [#permalink] New post   22 Apr 2010, 11:28
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Hi all!

How would you approach this kinda question:

How many divisors does 39690 have?
Correct Answer: 60


How do you attain the factors of any given number in general? Any simple offerings?



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Re: Divisors [#permalink] New post   22 Apr 2010, 11:37
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If a number can be expressed as

2^a x 3^b x 5^c x 7^d x 11^e.......

multiple of prime factors

then # of factors/divisors= (a+1)x(b+1)x(c+1)x(d+1)x(e+1)x.........

39690= 2^1 x3^4 x5^1x7^2
=> number of divisors = (1+1)x(4+1)x(1+1)x(2+1) = 2x5x2x3=10x6=60

Answer=60
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gmatt14
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Re: Divisors [#permalink] New post   22 Apr 2010, 11:43
Thx..........but how did you attain the prime numbers? I'm a lil bumped getting prime numbers out of huge numbers.
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Re: Divisors [#permalink] New post   22 Apr 2010, 11:44
gmatt14 wrote:
Hi all!

How would you approach this kinda question:

How many divisors does 39690 have?
Correct Answer: 60


How do you attain the factors of any given number in general? Any simple offerings?


try this:
https://gmatclub.com/forum/really-tough-number-theory-problem-92915.html

thanks
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Re: Divisors [#permalink] New post   22 Apr 2010, 12:02
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Conceptual Approach - Start by dividing the number by the prime numbers like 2,3,5,7,11 etc.... one by one...lengthy approach

Shortcut - You can apply divisibility tests, like 39690 is clearly divisible by 10 because units place is 0.So

39690=3969x10

Now we can clearly see that 3969 is divisible by 9 (because 3+9+6+9 = 27 which is divisible by 9)

39690=441x9x10

again 441 is divisible by 9 (4+4+1=9)

39690=49 x 9 x 9 x 10 = 7 x 7 x 3 x 3 x 3 x 3 x 5 x 2 = 2^1 x3^4 x5^1x7^2

Does that answer your question?
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gmatt14
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Re: Divisors [#permalink] New post   22 Apr 2010, 13:34
Thanks a lot. That helps. What if, such a number would have not been divisible by a multiple of 3?
Can you always find the factors by adding each digit?



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Re: Divisors [#permalink]
22 Apr 2010, 13:34
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