The goal of this problem is to find the exact 4 fractions such that

The fraction is less than one in value.
It contains two digits in the numerator and two digits in the denominator.
The numerator and denominator have a digit in common.
If you remove the common digit from both the numerator and the denominator, you'll get a fraction equal to what you get when you simplify the fraction in a proper way. For example, $49 / 98$ has $9$ as a common digit. If we remove it, then we'll get $4/8$ which is the actual value that you would get if you properly divide both the numerator and the denominator by $49$.
We should not count trivial examples. For example $30/50 = 3/5$ is a trivial example.

With just the first three conditions, the search space contains at most 1377 fractions. You can use the loop below to see them!

int count = 0;
for (int num = 10; num <= 99; num++) {
    for (int den = num+1; den <= 99; den++) {
        int n1 = num % 10;
        int n2 = num / 10;
        int d1 = den % 10;
        int d2 = den / 10;
        if (n1 == d1 || n1 == d2 || n2 == d1 || n2 == d2) {
            count++;
            printf("%d / %d with digits %d%d / %d%d \n", num, den, n2, n1, d2, d1);
        }
    }
}
printf("count = %d\n", count);

Solution

Since we have at most $1377$ cases, then we can easily check all of them. We will ignore trivial cases. For the non-trivial cases like $49/98$, we will extract the $4$ digits and save them in different variables. The numerator digits are $n_1 = 9$ and $n_2 = 4$. The denominator digits are $d_1 = 8$ and $d_2 = 8$. To check if $\frac{49}{98}$ is equal to $\frac{4}{8}$, we can check if

$$ \begin{align*} 49 \times 8 &= 98 * 4 \\ numerator \times d_1 &= denominator \times n_2 \end{align*} $$

After we collect all the fractions, we can divide by the greatest common divisor. This is captured in the following code.

int numerator = 1;
int denominator = 1;
for (int num = 10; num <= 99; num++) {
    for (int den = num+1; den <= 99; den++) {
        int n1 = num % 10;
        int n2 = num / 10;
        int d1 = den % 10;
        int d2 = den / 10;
        //
        if (n2 == d2 || n1 == d1) {
            continue; // even if it works, this is a trivial example
        }
        if (n1 == d2 || n2 == d1) {
            // example is 49/98. this will simplify to 4/8
            // now we want to check if 49/98 == 4/8 which is the same as checking if 49*8 == 98*4
            if (den * n2 == num * d1 || den * n1 == num * d2) {
                // printf("%d / %d\n", num, den);
                numerator *= num;
                denominator *= den;
            }
        }
    }
}
int gcd = std::gcd(numerator,denominator);
printf("denominator = %d\n", denominator/gcd);

The entire code is here.

References

Project Euler - 33