Answer to Question #269278 in Algebra for Random

Question

Find the smallest positive integer N that satisfies all of the following conditions:

 N is a square.

 N is a cube.

 N is an odd number.

 N is divisible by twelve different prime numbers.

How many digits does this number N have?

Answer 1

Let the number be N.

As N is odd, 2:is not a prime factor of it. For smallest value of N, first 12 prime factors excluding 2 will be prime factors of N.

Again exponent of every prime factor will be multiple of 2 as it is a square number.

Similarly exponent of every prime factor will be multiple of 3 as it is a cube number.

So exponent of every prime factor will be multiple of 6.

For least value of N, the exponent of every prime factor will be LCM of 2,3 i.e. 6.

So N = 3⁶.5⁶.7⁶.11⁶.13⁶.17⁶.19⁶.23⁶.29⁶.31⁶.37⁶.41⁶

= (3.5.7.11.13.17.19.23.29.31.37.41)⁶

So the smallest positive integer satisfying all the conditions is

N = (3.5.7.11.13.17.19.23.29.31.37.41)⁶

To find the number of digits in N we take logerithm of N with respect to base 10.

log₁₀(N) = log₁₀(3.5.7.11.13.17.19.23.29.31.37.41) ⁶

= 6log₁₀(3.5.7.11.13.17.19.23.29.31.37.41)

= 6*14.182201

= 85.093206

Characteristics of logerithm value is 85.

So number of digits in N is 86