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For a positive integer k written in decimal notation, #(k)

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kevincan
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For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   27 Sep 2006, 02:35
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For a positive integer k written in decimal notation, #(k) is defined as the product of the digits of k. For instance, #(333)=27 and #(142)=8. If Y is the set of four digit whole numbers w such that #(w)=60, what percent of the elements of Y are greater than 5000?



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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   27 Sep 2006, 03:11
25%

The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   27 Sep 2006, 22:38
i am getting 21/60

numbers could be permutation of 5322 or 1652 or 1453.
total numbers = !3/!2 + 2*3*2*1 + !3 = 21
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   27 Sep 2006, 23:02
10/21

Permutations of 4351 2651 , 2353 = 3 * 4!
Total possible combinations = 9P4 (we have to ignore 0)

3*4!/9P4 * 100 = 10/21
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   27 Sep 2006, 23:39
Solution please :o I thought 25% as well.
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   Updated on: 25 Sep 2008, 01:17
lanuk wrote:
25%

The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%


Hey lanuk,

it must be 3 of 12, 6 of 24 and 6 of 24.........so total 15 out of 60 ie 25%.

I too got the same answer. So let's wait for the OA

Originally posted by cicerone on 27 Sep 2006, 23:59.
Last edited by cicerone on 25 Sep 2008, 01:17, edited 1 time in total.
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   28 Sep 2006, 00:43
kevincan wrote:
cicerone wrote:
lanuk wrote:
25%

The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%


Hey lanuk,

it must be 3 of 12, [color=red]6 of 24 and 6 of 24.........so total 15 out of 60 ie 25%.

I too got the same answer. So let's wait for the OA[/color]


Why only 6 of 24?


Hey kevin i got it...............

it is 21/60 ie 35%
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Re: For a positive integer k written in decimal notation, #(k) [#permalink] For a positive integer k written in decimal notation, #(k)   28 Sep 2006, 02:12
Yup Cicerone..you're right..should be 21/60...
3 +12(Both starting with 5 and 6 are greater than 5000) +6
Silly mistake :oops:



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Re: For a positive integer k written in decimal notation, #(k) [#permalink]
28 Sep 2006, 02:12
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