Game Show

Since type $a$ has 1 item, and the next type must have at least twice as many, type $b$ can have the remaining $99$ items (since $99 \ge 2 \times 1$). Thus, the minimum number of different types of prizes is 2 ($a$ and $b$).

To maximize the number of types, we minimize the count of each type:

• $N_a = 1$
• $N_b = 2$
• $N_c = 4$
• $N_d = 8$
• $N_e = 16$
• $N_f = 32$

Sum for 6 types = $1 + 2 + 4 + 8 + 16 + 32 = 63 \le 100$. For 7 types, the minimum sum would be $63 + 64 = 127 > 100$, which is not possible. Thus, the maximum possible number of different types of prizes is 6.

If there are 45 items of type $c$ ($N_c = 45$), then $N_b \le 22.5 \implies N_b \le 22$, and $N_a \le 11$. But $N_c \ge 2 N_b \ge 4 N_a$. If $N_c = 45$, total items up to $c$ is $1 + 2 + 45 = 48$. If there is no type $d$, then total must be 100, but $1 + 2 + 45 = 48 \ne 100$. If there is type $d$, $N_d \ge 90$, but $48 + 90 = 138 > 100$. Thus, having exactly 45 items of type $c$ is not possible.

Total items of the type in box 45 = $31 + 1 + 43 = 75$. So one type has 75 items. Since $N_a=1$, and each subsequent type at least doubles the previous, the maximum number of preceding types before 75 can be tested:

• $N_a = 1$
• $N_b = 2$
• $N_c = 4$
• $N_d = 8$
• $N_e = 75$ Sum = $1 + 2 + 4 + 8 + 75 = 90 \le 100$. No 6th type $f$ is possible because $N_f \ge 2 \times 75 = 150 > 100$. Hence, the maximum number of types is 5.