IIT JEE , AIEEE Forum - Mathematics , Algebra

vikram1989

a) Given that a number is a perfect cube, what is the pobability that it is also a perfect square.(For Random numbers 1 to 1,000,000)

b) Given that a number is a perfect sqaure, what is the probability that it is also a perfect cube.(For Random numbers 1 to 1,000,000)

Working appriciated.

Thank you.

rudra.panda

1000,000=1000^2=100^3

hsbhatt

Every natural number N can be uniquely factorised into its prime factors:

$N = p_1^{\alpha_1} . p_2^{\alpha_2}...p_n^{\alpha_n} \ \text{where} \ p_1, p_2 \ \text{etc. are prime numbers and} \ \alpha_1, \alpha_2 ... \alpha_n \in \mathbb{N}$

If a number is a perfect square each of the prime factors must be raised to even powers

If a number is a perfect cube, each of the prine factors must be raised to powers that are multiples of 3.

Thus if a number is a perfect square and a perfect cube, each factors is raised to a power that is a multiple of 6 i.e. if N is such a number N = a⁶ where a is a natural number.

Since our upper bound is 10⁶, there are only 10 such numbers in the given range

for part (a) the number of perfect cubes is 100 of these only 10 are also perfect squares. Hence prob = 0.1

for part(b) the number of perfect squares is 1000 and so the prob = 0.01.

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