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# How many factors of 10800 are perfect squares?

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Retired Moderator
Status: The best is yet to come.....
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How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 06:44
1
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95% (hard)

Question Stats:

30% (02:08) correct 70% (01:50) wrong based on 159 sessions

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How many factors of 10800 are perfect squares?

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

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Hasan Mahmud
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Joined: 25 Feb 2013
Posts: 1196
Location: India
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How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 07:16
2
7
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

Option E
##### General Discussion
Math Expert
Joined: 02 Aug 2009
Posts: 7764
How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 06:53
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

total = 5+5+2=12 ways
E
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Re: How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 07:52
2
1
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.
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Re: How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 07:55
1
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

total = 5+5+2=12 ways
E

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

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20 Oct 2017, 11:18
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
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Posts: 7764
Re: How many factors of 10800 are perfect squares?  [#permalink]

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20 Oct 2017, 21:07
niks18 wrote:
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

total = 5+5+2=12 ways
E

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.

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

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09 Jan 2018, 13:31
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)
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Re: How many factors of 10800 are perfect squares?  [#permalink]

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13 Jan 2019, 22:44
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Re: How many factors of 10800 are perfect squares?   [#permalink] 13 Jan 2019, 22:44
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