**Problem Statement:**

The product of two numbers is 234, and their highest common factor (HCF) is 13. We need to find the number of such pairs.

**Solution:**

To solve this problem, we will use the concept of prime factorization and the relationship between the product of two numbers and their highest common factor.

**Step 1: Prime Factorization of 234**

To find the prime factorization of 234, we divide it successively by prime numbers until we reach 1.

234 divided by 2 gives us 117.
117 divided by 3 gives us 39.
39 divided by 3 gives us 13.
13 divided by 13 gives us 1.

So, the prime factorization of 234 is 2 × 3 × 3 × 13.

**Step 2: Relationship between Product and HCF**

Let the two numbers be A and B. We know that the product of two numbers is equal to the product of their prime factors raised to their respective powers.

A × B = 2^a × 3^b × 13^c

Also, the highest common factor (HCF) of two numbers is equal to the product of the common prime factors raised to the minimum of their respective powers.

HCF(A, B) = 2^x × 3^y × 13^z

In this case, the HCF is 13, so x = 0, y = 0, and z = 1.

**Step 3: Finding the Number of Pairs**

Now, we need to find the number of pairs (A, B) that satisfy the given conditions.

From the above relationship, we can conclude that the prime factorization of A and B should be of the form:

A = 2^p × 3^q × 13^1
B = 2^r × 3^s × 13^1

The product of A and B is given by:

A × B = 2^(p+r) × 3^(q+s) × 13^2

From the given information, we have:

2^(p+r) × 3^(q+s) × 13^2 = 2 × 3 × 3 × 13

Comparing the exponents of the prime factors on both sides, we get:

p + r = 1
q + s = 1

The possible values of (p, r) and (q, s) are:

(0, 1) and (1, 0)

Therefore, there are two possible pairs (A, B) that satisfy the given conditions:

A = 2^0 × 3^1 × 13^1 = 3 × 13 = 39
B = 2^1 × 3^0 × 13^1 = 2 × 13 = 26

and

A = 2^1 × 3^0 × 13^1 = 2 × 13 = 26
B = 2^0 × 3^1 × 13^1 = 3 × 13 = 39

Hence, there are two such pairs (39, 26) and (26, 39) that satisfy the given conditions.