Question

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

Answer 1

Let's consider a number with prime factorization pa⋅qb, where p and q are distinct prime numbers, and a and b are positive integers. The number of factors (N) of this number is given by (a+1)⋅(b+1).

For a number with 15 factors, we can have either a=14 and b=0 or a=2 and b=4 (or vice versa).

1. If a=14 and b=0, the number is p14.
2. If a=2 and b=4, the number is 2⋅p2⋅q4.

Now, we need to find the two smallest natural numbers with these factorizations.

1. For a=14 and b=0, the smallest number is 214=16,384214=16,384.
2. For a=2 and b=4, the smallest number is 22⋅34=14422⋅34=144.

The sum of the two smallest natural numbers with 15 factors is 16,384+144=16,52816,384+144=16,528. 

Therefore, the answer is 16,528.

Answer 2

We know that the number of factors of these two numbers is \(15\).
Factors of \(15 =1, 3, 5,15\)
The number of factors of \(N\) = \((p + 1) (q + 1)\)
Where,
\(N = a^P \times b^q\) and \(a\), \(b\) are prime numbers. 
Therefore, the value of \(N\) will be least when \((p+1)\) and \((q+1)\) are as close as possible, and a and b are the smallest distinct prime numbers.
Hence,
\(p+1 = 3\)
\(p = 3-1=2\)
and.
\(q+1 = 5\)
\(q = 5-1=4\),
The prime numbers \(a =1\) and \(b=2\)
Thus, the lowest value of N,
\(= 2^4\times 3^2\)
\(= 144\)
The second lowest value of N,
\(= 2^2 \times 3^4\)
\(= 324\)
Now, sum \(= 144+324 = 468 \)

So, the answer is \(468\).