__Perfect cube factors:__

If a number is a perfect cube, then the power of the prime factors should be divisible by 3.

**Example 1:Find the number of factors of2 ^{9}3^{6}5^{5}11^{8} that are perfect cube?**

**Solution:**If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2

^{(0 or 3 or 6or 9)}—– 4 factors

3

^{(0 or 3 or 6)}Â —–Â 3Â factors

5

^{(0 or 3)}——- 2 factors

11

^{(0 or 3 or 6 )}— 3 factors

Hence, the total number of factors which are perfect cube 4x3x2x3=72

__Perfect square and perfect cube__If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.

**Example 2**:

**How many factors of 2**

^{9}3^{6}5^{5}11^{8}are both perfect square and perfect cube?**Solution:**If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2

^{(0 or 6)}—– 2 factors

3

^{(0 or 6)}—– 2 factors

5

^{(0)}——- 1 factor

11

^{(0 or 6)}— 2 factors

Hence total number of such factors are 2x2x1x2=8

**Example 3: How many factors of2**

^{9}3^{6}5^{5}11^{8}are either perfect squares or perfect cubes but not both?**Solution:**

Let A denotes set of numbers, which are perfect squares.

If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have

2

^{(0 or 2 or 4 or 6 or 8)}—– 5 factors

3

^{(0 or 2 or 4 or 6)}Â —– 4 factors

5

^{(0 or 2or 4 )}——- 3 factors

11

^{(0 or 2or 4 or6 or 8 )}— 5 factors

Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300

Let B denotes set of numbers, which are perfect cubes

If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2

^{(0 or 3 or 6or 9)}—– 4 factors

3

^{(0 or 3 or 6)}Â —–Â 3Â factors

5

^{(0 or 3)}——- 2 factors

11

^{(0 or 3 or 6 )}— 3 factors

Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2

^{(0 or 6)}—– 2 factors

3

^{(0 or 6)}—– 2 factors

5

^{(0)}——- 1 factor

11

^{(0 or 6)}— 2 factors

Hence total number of such factors are i.e.n(Aâˆ©B)=2x2x1x2=8

We are asked to calculate which are either perfect square or perfect cubes i.e.

n(A U B )= n(A) + n(B) – n(Aâˆ©B)

=300+72 â€“ 8

=364

Hence required number of factors is 364.