Question

let A be the least number such that 10A is a perfect square and 35A is a perfect cube . then the number of positive divisors of A is

Answer 1

Answer:

72

Step-by-step explanation:

Let A be the least number such that 10A is a perfect square and 35A is a perfect cube . then the number of positive divisors of A is

10A is a perfect Square

if A  is 10b²

then 10A  would be 10(10b²)  = (10b)²

35A is a Perfect cube

if A is 35²c³

then 35A would be   35(35²c³)  = (35c)³

10b² = 35² c³

=> 10b² = (35c)²c

=> c = 10

Hence

10b² = 35² * 10³

= (5 * 7)² * (2 * 5)³

= 2³ * 5⁵ * 7²

= 1225000

Number of Factors = (3 + 1)(5 + 1)(2+1)

= 4 * 6 * 3

= 72