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# divisors Psq

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Manager
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28 May 2009, 09:57
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how many divisors of 21600 are perfect squares
Manager
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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
Manager
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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.

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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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Manager
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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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rampuria

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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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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 04 Sep 2006
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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
Manager
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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 ?
Senior Manager
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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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Lahoosaher

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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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Intern
Joined: 12 May 2009
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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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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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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: divisors Psq   [#permalink] 01 Jun 2009, 15:27
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