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This is a fairly straightforward question on the topic of Factors & Multiples. So, the best approach is to prime factorise the given number and then use the formula.

If N is a composite number which can be written as N = \(a^p * b^q * c^r *\)…………, then the number of factors of N = (p+1) * (q+1) * (r+1)*……

1764 = \(2^2 * 3^2 * 7^2\).

From this we gather that 1764 itself is a perfect square. Therefore, all the prime factors of 1764 have even powers. Also, the total number of factors = (2+1) * (2+1) * (2+1) = 3*3*3 = 27.

Now, how do we find out how many of these are perfect squares? Let’s look at the diagram below:

Attachment:
25th June 2019 - Reply 2.JPG
25th June 2019 - Reply 2.JPG [ 35 KiB | Viewed 5798 times ]

Under each prime factor, you see that there are 2 options listed out. Since there are 3 prime factors, we can form a total of 8 combinations.

For example, \(2^0 * 3^0*7^0\) = 1 is a factor of 1764 and is a perfect square. Similarly, \(2^0 * 3^2 * 7^0\) is also a factor of 1764 and a perfect square.
This way, we can have 8 such products which are all factors of 1764 and are perfect squares. Therefore, the correct answer has to be option C.

Hope this helps!
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The easiest way I found but a little time consuming was just take the LCM of the number:

\(1764=2^2*3^2*7^2\)

Then as all the factors (2,3 and 7) are perfect squares, we play with the factors:

\(2^2 --- 1\)
\(3^2 --- 2\)
\(7^2 --- 3\)
\(2^2*3^2*7^2\) All together (2*3*7)^2 --- 4
\(2^2*3^2 --- 5\)
\(2^2*7^2 --- 6\)
\(3^2*7^2 --- 7\)

And finally do not forget 1 --- 8 factors in total.
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Dillesh4096
How many factors of 1764 are perfect squares ?
A. 4
B. 7
C. 8
D. 9
E. 10

Posted from my mobile device

Asked: How many factors of 1764 are perfect squares ?

1764=2^2*3^2*7^2

Perfect squares are made when power of 2,3 and 7 are either 0 or 2.

Total ways = 2*2*2=8

IMO C
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CrackVerbalGMAT
This is a fairly straightforward question on the topic of Factors & Multiples. So, the best approach is to prime factorise the given number and then use the formula.

If N is a composite number which can be written as N = \(a^p * b^q * c^r *\)…………, then the number of factors of N = (p+1) * (q+1) * (r+1)*……

1764 = \(2^2 * 3^2 * 7^2\).

From this we gather that 1764 itself is a perfect square. Therefore, all the prime factors of 1764 have even powers. Also, the total number of factors = (2+1) * (2+1) * (2+1) = 3*3*3 = 27.

Now, how do we find out how many of these are perfect squares? Let’s look at the diagram below:

Attachment:
25th June 2019 - Reply 2.JPG

Under each prime factor, you see that there are 2 options listed out. Since there are 3 prime factors, we can form a total of 8 combinations.

For example, \(2^0 * 3^0*7^0\) = 1 is a factor of 1764 and is a perfect square. Similarly, \(2^0 * 3^2 * 7^0\) is also a factor of 1764 and a perfect square.
This way, we can have 8 such products which are all factors of 1764 and are perfect squares. Therefore, the correct answer has to be option C.

Hope this helps!

Can we simply multiply the even powers? Or is it necessary to list the powers in 0?
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