How many factors of 1764 are perfect squares ? A. 4 B. 7 C.

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:



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!

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.

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

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