South African Mathematical Olympiad, Third Round

We are looking at sums of the form
$$s(10 + \ell, k) = \sum_{i=0}^{k-1} (10 + \ell + i) = k\left(10 + \ell + \frac{1}{2}(k-1)\right),$$
where $\ell \ge 0$ ($10 + \ell$ represents the starting number of the consecutive numbers in a set) and $k \ge 2$ represents the number of consecutive numbers in a set. Each of these sums (the $(k+1)$-th number in the set) must be less than, or equal to, $99$. By taking $\ell = 0$, we see that the maximum possible value for $k$ is $7$.

For the case $k = 2$ (i.e., where three cards are used), we must have $2 \cdot (10+\ell+\frac{1}{2}(2-1)) \le 99$, so that $\ell \le 39$. The $39+1=40$ possibilities of three cards are: $10+11=21, 11+12=23, \dots, 49+50=99$.

For the case $k = 3$ (i.e., where four cards are used), we get $3 \cdot (10+\ell+\frac{1}{2}(3-1)) \le 99$, so that $\ell \le 22$, representing the $22+1=23$ groups of four cards: $10+11+12=33, \dots, 32+33+34=99$.

Continuing in this way, we find that $\ell \le 13$ when $k = 4$, $\ell \le 7$ when $k = 5$, $\ell \le 4$ when $k = 6$, and $\ell \le 1$ when $k = 7$. All in all, there are $40+23+14+8+5+2=92$ sets of cards satisfying the conditions.