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Bunuel
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Geet03
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twobagels
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it's 8, unless, i missed some, which is definitely possible.

225, 100, 36, 25, 9, 16, 4, 900
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Rajat8
How many divisors of 21,600 are perfect squares?

Solution: Let's do prime factorization of 21,600 i.e. \(21,600 = 2^5 * 3^3 * 5^2\)

For a number to be a square the exponent of the prime ALL prime factors should be an even number.
Thus, from the above prime factorization \(2^2 , 3^2, 2^4, 5^2\) are some of the perfect squares. The combinations of these 4 squares are also perfect squares:

1. \(2^2 = 4\)
2. \(3^2 = 9\)
3. \(2^4 = 16\)
4. \(5^2 = 25\)
5. \(2^2 * 3^2 = 36\)
6. \(2^2 * 5^2 = 100\)
7. \(3^2 * 2^4 = 144\)
8. \(3^2 * 5^2 = 225\)
9. \(2^4 * 5^2 = 400\)
10. \(2^2 * 3^2 * 5^2 = 900\)
11. \(3^2 * 2^4 * 5^2 = 3600\)

Answer should be 11 divisors.

You have missed 1. so total will be 12
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Let's prime factorize the number: 21600.
A simple way to do so just to keep 21600 as 216 x 100

Now,
Prime Factorization of 216 = (2^3) x (3^3)
Prime Factorization of 100 = (2^2) x (5^2)
Prime Factorization of 21600 = (216 x 100) = (2^3) x (3^3) x (2^2) x (5^2) = (2^5) x (3^3) x (5^2)

Concept: A perfect square will definitely contain even the power of the number (which is prime of course). So,

21600 = (2^0, 2^1, 2^2, 2^3, 2^4, 2^5) x (3^0, 3^1, 3^2, 3^3) x (5^0, 5^1, 5^2, 5^3)

All the power contain even numbers including 0 will be considered, i.e.,

21600 = (2^0, 2^1, 2^2, 2^3, 2^4, 2^5) x (3^0, 3^1, 3^2, 3^3) x (5^0, 5^1, 5^2, 5^3)

So, No. of perfect squares of 21600 = 3 x 2 x 2 = 12 (Ans)
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