Question

Find the sum of the two smallest natural numbers with the number of factors as 15.

Answer 1

Consider a number with prime factorization \(p^a \cdot q^b\), where \(p\) and \(q\) are distinct prime numbers, and \(a\) and \(b\) are positive integers. The count of factors for this number is \((a+1)(b+1)\).

For a number with 15 factors, possible pairs of \((a, b)\) (or \((b, a)\)) are \((14, 0)\) or \((2, 4)\).

1. If \(a=14\) and \(b=0\), the number's form is \(p^{14}\).
2. If \(a=2\) and \(b=4\), the number's form is \(p^2 \cdot q^4\).

We need to determine the two smallest natural numbers exhibiting these factorizations.

1. For the form \(p^{14}\), the smallest number is \(2^{14} = 16,384\).
2. For the form \(p^2 \cdot q^4\), the smallest number is \(2^2 \cdot 3^4 = 144\).

The sum of these two smallest natural numbers with 15 factors is \(16,384 + 144 = 16,528\).

The result is 16,528.