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divisors Psq

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divisors Psq [#permalink]

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New post 28 May 2009, 09:57
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how many divisors of 21600 are perfect squares

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Re: divisors Psq [#permalink]

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New post 28 May 2009, 14:37
I believe its 5

21600

216 * 100 1

54 * 4 * 100 2

9 * 5 * 4 * 100 3

also

100 = 25 * 4 4

and 4*4=16 5

not sure if there are any others tho

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New post 29 May 2009, 20:34
Prime factorization of 21600 = \(2^5\) \(3^3\) \(5^2\) = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5

Since, the only prime factors for 21600 are 2,3,5 no other prime numbers are factors for it.

Now, form a symmetrical pattern of numbers by choosing 2, 4, 6, 8 or 10 of these numbers.

(2) x (2) = 4

(2x2) x (2x2) = 16

(3) x (3) = 9

(5) x (5) = 25

(3x2) x (3x2) = 36

(5x2) x (5x2) = 625

(2x3x5) x (2x3x5) = 900

(2x2x3x5) x (2x2x3x5) = 3600

We have to include 1 as well because it is a perfect square in itself.


Thus, answer is 9.

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New post 29 May 2009, 22:28
vcbabu wrote:
how many divisors of 21600 are perfect squares


21600 = (2^5 x 3^3 x 5^2) = \((2^2)\) \((2^2)\) \((2)\) \((3^2)\) \((3)\) \((5^2)\)

1 x 1 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
10^2 = 100
12^2 = 144
15^2 = 225
20^2 = 400
30^2 = 900
60^2 = 3600

Total = 12 if I am not missing anything....

Edited - for 21,600 as I misread 21,600 as 216,000.
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Re: divisors Psq [#permalink]

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New post 30 May 2009, 00:04
IMO 10

\(21600 = 2^5 X 3^3 X 5^2 X 1^2\)

This gives us \(2^4 , 2^2, 3^2, 5^2, 1^2\) as divisors which are perfect squares

if you work out with the above you will come to 10
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Re: divisors Psq [#permalink]

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New post 30 May 2009, 01:07
I believe that this can be done very systematically as follows.

21600 = 2^5 * 3^2 * 5^2

Now let identify those terms that are squares:

21600 = 2 * 4^2 * 9^1 * 25^1

We can ignore the "2" (we only care about the factors that are perfect squares) and this becomes a simple combination problem.

There are 3 possible exponents of 4 (0, 1 and 2), 2 of 9 (0 and 1) and 2 of 25 (0 and 1)

Hence there are 3*2*2 = 12 possible combination of square divisors:

{1, 4, 9, 16, 25, 36, 100, 144, 225, 400, 900, 3600}
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Re: divisors Psq [#permalink]

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New post 31 May 2009, 00:30
21600 = 2^5*3^3*5^2

all the divisors made by even powers of 2,3,5 will be perfect squares.

even powers of 2 : 2^0 , 2^2, 2^2 tot 3

even powers of 3: 3^0 , 3^2 tot 2

even powers of 5 :5^0 , 5^2 tot 2


tot : 3*2*2 =12

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New post 31 May 2009, 06:28
vcbabu wrote:
21600 = 2^5*3^3*5^2

all the divisors made by even powers of 2,3,5 will be perfect squares.

even powers of 2 : 2^0 , 2^2, 2^2 tot 3

even powers of 3: 3^0 , 3^2 tot 2

even powers of 5 :5^0 , 5^2 tot 2


tot : 3*2*2 =12


I think you meant \(2^4\). right ?

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Re: divisors Psq [#permalink]

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New post 31 May 2009, 16:23
21600 = 2^5 * 3^2 * 5^2

2^0
1-1
2-(3) x (3) = 9
3-(5) x (5) = 25
4-(3x5) x (3x5)=625

2^2
5-(2x2) = 4
6-(2x2)(2x2)
7-(2x2)x(3x3)=36
8-(2x2)x(5x5)=100
9-(2x2)x(3x3)x(5x5)=900

2^4

10-(2x2)x(2x2)x(3x3)=144
11-(2x2)x(2x2)x(5x5)=400
12-(2x2)x(2x2)x(3x3)x(5x5)=3600
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New post 31 May 2009, 17:23
Quote:
how many divisors of 21600 are perfect squares


21600=2*2*2*2*2*3*3*3*5*5

2^2
3^3
5^5
2x2^2x2
2x3^2x3
2x5^2x5
3x5^3x5
2x2x3^2x2x3
2x2x5^2x2x5
2x3x5^2x3x5

OA = 10
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Re: divisors Psq [#permalink]

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New post 31 May 2009, 22:26
21600 = 2^5 * 3^2 * 5 ^ 2
=> 2 * 2^4 * 3^2 * 5^2 --> this gives us the perfect square combination
=> 4^2 * 9^1 * 25^1
possible values for substitution = ((0,1,2),(0,1),(0,1))
=> 3*2*2 = 12

IMO 12

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Re: divisors Psq [#permalink]

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New post 01 Jun 2009, 15:27
vannu wrote:
Quote:
how many divisors of 21600 are perfect squares


21600=2*2*2*2*2*3*3*3*5*5

2^2
3^3
5^5
2x2^2x2
2x3^2x3
2x5^2x5
3x5^3x5
2x2x3^2x2x3
2x2x5^2x2x5
2x3x5^2x3x5

OA = 10


You are missing 1 and (2x2x3x5)x(2x2x3x5) for a total of 12.
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Re: divisors Psq   [#permalink] 01 Jun 2009, 15:27
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