How many factors of 10800 are perfect squares?

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.

Hi chetan2u

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.

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

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

total 12

Prime factorization: \(10800\) = \(2^4 * 5^2 * 3^3\)
= \(4^2 * 25^1 * 9^1 * 3\)

To get factors which are perfect squares : we need to consider factors from : \(4^2 * 25^1 * 9^1\)

set A: { \(4^0, 4^1, 4^2\) } = 3 (total)
set B: { \(9^0, 9^1\) } = 2 (total)
set C: { \(25^0, 25^1\) } = 2 (total)

Total combinations : 3 * 2 * 2 (picking anyone from each set) = 12 => (E)

Why do we multiply the different types of combinations instead of summing them? Thanks

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