The number of positive integers less than $50$, having exactly two distinct factors other than $1$ and itself, is

We need to find numbers that have exactly two distinct factors other than 1 and itself.

If a number has exactly two distinct factors other than 1 and itself, then the total number of factors is:

1 (always a factor)

The number itself (always a factor)

2 additional distinct factors

Total factors = 4

So we're looking for numbers with exactly 4 factors.

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From number theory, there are only two types of numbers that have exactly 4 factors:

Numbers of the form $p^3$ (prime cubed):

When we have $p^3$ where $p$ is prime:

Factors are: $1, p, p^2, p^3$

Count: exactly 4 factors

Numbers of the form $p \times q$ (product of two distinct primes):

When we have $p \times q$ where $p$ and $q$ are different primes:

Factors are: $1, p, q, p \times q$

Count: exactly 4 factors

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Case 1: Numbers of the form $p^3$

Prime cubes less than 50:

$2^3 = 8 < 50$

$3^3 = 27 < 50$

$4^3 = 64 > 50$ (and 4 isn't prime anyway)

Count from Case 1: 2 numbers

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Case 2: Numbers of the form $p \times q$

Starting with 2 as the smaller prime:

$2 \times 3 = 6$

$2 \times 5 = 10$

$2 \times 7 = 14$

$2 \times 11 = 22$

$2 \times 13 = 26$

$2 \times 17 = 34$

$2 \times 19 = 38$

$2 \times 23 = 46$

Starting with 3 as the smaller prime:

$3 \times 5 = 15$

$3 \times 7 = 21$

$3 \times 11 = 33$

$3 \times 13 = 39$

$3 \times 17 = 51 > 50$

Starting with 5 as the smaller prime:

$5 \times 7 = 35$

$5 \times 11 = 55 > 50$

Count from Case 2: 13 numbers

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Total numbers = Numbers from Case 1 + Numbers from Case 2

Total numbers = $2 + 13 = 15$

The answer is 15.