Question

The positive integer X is divisible by both 9 and 21. What is the smallest possible number of positive integers that will divide evenly into x?

Answer 1

To find the smallest positive integer \( X \) that is divisible by both 9 and 21, we need to determine the least common multiple (LCM) of these two numbers.

First, we find the prime factorization of both numbers:
- The prime factorization of 9 is \( 3^2 \).
- The prime factorization of 21 is \( 3^1 \times 7^1 \).

To find the LCM, we take the highest power of each prime factor that appears in the factorizations:

- For the prime \( 3 \): The maximum power is \( 3^2 \) (from 9).
- For the prime \( 7 \): The maximum power is \( 7^1 \) (from 21).

Thus, we can calculate the LCM as follows:

\[
\text{LCM}(9, 21) = 3^2 \times 7^1 = 9 \times 7 = 63.
\]

Now that we have determined \( X = 63 \), we need to find the number of positive integers that divide evenly into 63.

To do this, we first perform the prime factorization of 63:

\[
63 = 3^2 \times 7^1.
\]

Next, to find the number of divisors, we use the formula for calculating the number of divisors based on the prime factorization. If a number \( n \) has the prime factorization \( p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_k^{e_k} \), then the number of divisors \( d(n) \) is given by:

\[
d(n) = (e_1 + 1)(e_2 + 1) \ldots (e_k + 1).
\]

For \( 63 = 3^2 \times 7^1 \), we have:

- The exponent of \( 3 \) is \( 2 \), so \( e_1 + 1 = 2 + 1 = 3 \).
- The exponent of \( 7 \) is \( 1 \), so \( e_2 + 1 = 1 + 1 = 2 \).

Therefore, the number of positive divisors of 63 is:

\[
d(63) = (2 + 1)(1 + 1) = 3 \times 2 = 6.
\]

Thus, the smallest possible number of positive integers that will divide evenly into \( x \) is

\[
\boxed{6}.
\]