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Approximately what percent of the positive factors of 1800

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Approximately what percent of the positive factors of 1800 [#permalink] New post 07 Jul 2006, 22:57
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Approximately what percent of the positive factors of 1800 are less than 30?

(A) 22% (B) 28% (C) 32% (D) 38% (E) 42%

Last edited by kevincan on 07 Jul 2006, 23:42, edited 2 times in total.
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 [#permalink] New post 07 Jul 2006, 23:27
E = 43%

1800 = (2*2*2)*(3*3)*(5*5)

consider a = 2, b = 3 and c =5

Factors are = 1 (No 1) + 3(Single a,b,c) + 3(Double aa,bb,cc) + 3(Double ab,bc,ca)+ 7 (Triplets aaa,aab,aac,bba,bbc,cca,ccb) + 8(Quartets aaab, aaac, aabb, aacc, bbcc, abcc, abbc,aabc) + 5 (Pair of five aaabb, aaacc, aabbc, aabcc, abbcc) + 3 (Pair of six aaabbc, aaabcc, aabbcc) + 1 (1800)

Total factors = 34

Factors less than 30 = 15

Answer = 43%

Is there a shorter way???
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 [#permalink] New post 08 Jul 2006, 00:50
1800=3^2*2^3*5^2 or 36 factors in total.
Less than 30 are 2,3,5,6,15,10,18,12,24,20,9,4,1,8 or 14 in total . Then 14/36=7/18 or 38%
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 [#permalink] New post 08 Jul 2006, 02:30
BG wrote:
1800=3^2*2^3*5^2 or 36 factors in total.
Less than 30 are 2,3,5,6,15,10,18,12,24,20,9,4,1,8 or 14 in total . Then 14/36=7/18 or 38%



Kevin,
can you please the answer as well as the shortest way to do this ?

thanks..
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 [#permalink] New post 08 Jul 2006, 05:50
When you look at the prime factorization of 1800, i.e. 2^3*3^2*5*2,

you can see that every factor of 1800 is of the form 2^x*3^y*5*z

where x is one of {0,1,2,3}, y is one of {0,1,2} and z is one of {0,1,2}

So, 3600 has 4*3*3=36 factors

How many are less than 30?

Look at possibilities for 3^y*5^z. These are 1,3,5,9,15,25

x=0 generates factors of 1, 3, 5, 9,15 ,25 keep all 6
x=1 " " 2, 6, 10,18 keep only 4
x=2 " " 6, 12,20, - keep 3
x=3 " " 12,24,-,- keep 2

So there are 15 factors out of 36 approx 42%

Last edited by kevincan on 09 Jul 2006, 01:00, edited 2 times in total.
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 [#permalink] New post 08 Jul 2006, 19:26
kevincan wrote:
When you look at the prime factorization of 1800, i.e. 2^3*3^2*5*2,

you can see that every factor of 1800 is of the form 2^x*3^y*5*z

where x is one of {0,1,2,3}, y is one of {0,1,2} and z is one of {0,1,2}

So, 3600 has 4*3*3=36 factors

How many are less than 30?

Look at possibilities for 3^y*5^z. These are 1,3,5,15

x=0 generates factors of 1, 3, 5, 15 keep all 4
x=1 " " 2, 6, 10,30 keep only 3
x=2 " " 6, 12,20, - keep 3
x=3 " " 12,24,-,- keep 2

So there are 14 factors out of 36 7/18 approx 38-39%


Factors less than 30 are 15 not 14.

1,2,3,4,5,6,8,9,10,12,15,18,20,24,25

15/36 = 41.67% which is closest to 43% not to 38%.
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 [#permalink] New post 09 Jul 2006, 06:10
this one took a long time but got close to 42%...phew
  [#permalink] 09 Jul 2006, 06:10
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