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Factors of \(10800=1*2^5*3^3*5^2\) all prime factors have atleast power of 2

so ways.. 1) single digits.. 1,2,3,4,5.... so 5 of them 2) product of two prime factors.. 2*3 2*5 3*5 3*4 4*5 so 5 ways 3) product of 3 prime factors 2*3*5 3*4*5 so 2 ways

How many factors of 10800 are perfect squares? [#permalink]

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20 Oct 2017, 06:16

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Mahmud6 wrote:

How many factors of 10800 are perfect squares?

A. 4 B. 6 C. 8 D. 10 E. 12

\(10800=1*2^4*3^3*5^2\)

For a factor to be a square it needs to have an even number of powers of each of the prime factors, in this case for \(2\), \(3\) & \(5\)

so for the sake of explanation, let \(10800=2^a*3^b*5^c\)

Now \(a\) can take values \(0\), \(2\) & \(4\) i.e \(3\) values

\(b\) can take values \(0\) & \(2\) i.e. \(2\) values

and \(c\) can take values \(0\) & \(2\) i.e. \(2\) values

Hence total number of perfect squares \(= 3*2*2=12\)

Again just to explain why we need to multiply here - All the square factors occur when we take combinations of exponents from the three sets - {0,2,4}, {0,2} & {0,2}. hence the rule of multiplication is applied here for counting

Re: How many factors of 10800 are perfect squares? [#permalink]

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20 Oct 2017, 06:52

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E. 12

10800 = \(3^3 * 2^4 * 5^2\)

For a factor to be perfect square it needs to have even powers of 3, 2, 5. Hence we count the number of even exponents of 3, 2, 5 and multiply them (combination activity)

The exponent count would be 0 and 2 for 3 - (total of 2) 0, 2, and 4 for 2 (total of 3) 0 and 2 for 5 (total of 2)

Total number of perfect square factor is 2 x 3 x 2 = 12 Why zero is included? because 1 (for example \(2^0 * 5^0\) = 1 x 1) is also a factor and it is a square.

Take an easy number 36 - How many factors of 36 are perfect squares? 36 = \(3^2 * 2^2\) Even exponents - 0, 2 for 2 and 0, 2 for 3 Total exponent count is two each for 2 and 3 Hence perfect square factors are 2 x 2 = 4

Long method Factors of 36 = 1, 2, 3, 4, 6, 8, 12, 36 Perfect sq factors are 1, 4, 9 and 36 - Hence 4 factors are perfect squares.

Re: How many factors of 10800 are perfect squares? [#permalink]

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20 Oct 2017, 06:55

chetan2u wrote:

Mahmud6 wrote:

How many factors of 10800 are perfect squares?

A. 4 B. 6 C. 8 D. 10 E. 12

hi..

Factors of \(10800=1*2^5*3^3*5^2\) all prime factors have atleast power of 2

so ways.. 1) single digits.. 1,2,3,4,5.... so 5 of them 2) product of two prime factors.. 2*3 2*5 3*5 3*4 4*5 so 5 ways 3) product of 3 prime factors 2*3*5 3*4*5 so 2 ways

You have got the right answer but I am not able to understand your approach. can you explain it in more detail. Also I don't think we can use summation for number of ways here.

Re: How many factors of 10800 are perfect squares? [#permalink]

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20 Oct 2017, 10:18

Another way:

10800 = 2^4 * 5^2 * 3^3

Perfect squares are those where all powers are even (multiples of 2). First step thus is to take out something common and leave in parenthesis everything that is a perfect square.

So 10800 = 3 * (2^4 * 5^2 * 3^2). Now whatever is inside parenthesis is a perfect square. Lets see what is it the square of ?

We can re-write this as: 3 * (2^2 * 5 * 3)^2. Note that the expression after 3 is whole square of (2^2 * 5 * 3). Now how many factors does 2^2 * 5 * 3 have? 3*2*2 = 12.

That's our answer. (basically whatever factor you take out of 2^2 * 5 * 3 will give you a new perfect square because its already raised to a power of 2)

This given number 10800 has 12 such factors which are perfect squares. Hence E answer

Factors of \(10800=1*2^5*3^3*5^2\) all prime factors have atleast power of 2

so ways.. 1) single digits.. 1,2,3,4,5.... so 5 of them 2) product of two prime factors.. 2*3 2*5 3*5 3*4 4*5 so 5 ways 3) product of 3 prime factors 2*3*5 3*4*5 so 2 ways

You have got the right answer but I am not able to understand your approach. can you explain it in more detail. Also I don't think we can use summation for number of ways here.

Hi...

the first step has been to factorize 10800, which is 2^5*3^3*5^2... here all are atleast to POWER of 2 and 2^5 also includes 4^2

what is left is to find different combinations or choosing 1 or more out of prime factors - 1,2,3,2^2,5 single - 1,2,3,4,5 THAT is it contains \(1^2,2^2,3^2,4^2,5^2\) two at a time - 2*3,2*5,3*5,3*4,4*5, THAT is it contains \((2*3)^2, (2*5)^2.......\) three at a time - 2*3*5, 3*4*5 THAT is it contains \((2*3*5)^2\) and \((5*4*3)^2\)