Question:
How many factors of N=26 x 38 x 510 x 712 are perfect squares as well as perfect cubes?
How many factors of N=26 x 38 x 510 x 712 are perfect squares as well as perfect cubes?
Updated On: Sep 13, 2024
- 28
- 20
- 22
- 24
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The Correct Option is D
Solution and Explanation
The correct option is (D): 24.
Perfect squares are even powers of all their prime factors while perfect cubes have powers which are multiples of 3 of all their prime factors. For the number to be both a perfect square as well as a perfect cube, then all the powers of its prime factors should be multiple of 6 (LCM of 2 and 3).
Consider a factor of the number, that will be a perfect square as well as a perfect cube = 2a x 3b x 5c x 7d
Now, ’a’ can be 0 or 6{2 possibilities}. ’b’ can be 0 or 6 {2 possibilities}.
‘c’ can be 0 or 6 {2 possibilities} and ‘d’ can be 0 or 6 or 12 {3 possibilities}
Thus, required possibilities will be 2 × 2 × 2 × 3 = 24.
Perfect squares are even powers of all their prime factors while perfect cubes have powers which are multiples of 3 of all their prime factors. For the number to be both a perfect square as well as a perfect cube, then all the powers of its prime factors should be multiple of 6 (LCM of 2 and 3).
Consider a factor of the number, that will be a perfect square as well as a perfect cube = 2a x 3b x 5c x 7d
Now, ’a’ can be 0 or 6{2 possibilities}. ’b’ can be 0 or 6 {2 possibilities}.
‘c’ can be 0 or 6 {2 possibilities} and ‘d’ can be 0 or 6 or 12 {3 possibilities}
Thus, required possibilities will be 2 × 2 × 2 × 3 = 24.
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