AIME 2026 II · 第 5 题

Solution 1

Let $r$ be the number of red marbles and $b$ be the number of blue marbles in the urn. The total number of marbles is $n = r + b$, with the constraints

$$r \geq 7,\quad b \geq 7.$$ Since the total number of ways to choose 7 marbles is $\dbinom{n}{7}$ for both events, their favorable outcomes satisfy:

$$\dbinom{r}{4}\dbinom{b}{3}=\dbinom{r}{5}\dbinom{b}{2}.$$ We simplify the equation using the combinatorial definition $\dbinom{k}{m}=\dfrac{k!}{m!(k-m)!}$:

$$\frac{r!}{4!(r-4)!}\cdot\frac{b!}{3!(b-3)!}=\frac{r!}{5!(r-5)!}\cdot\frac{b!}{2!(b-2)!}.$$ We cancel common terms and reduce factorials:

$$\frac{b-2}{3}=\frac{r-4}{5}.$$ We cross-multiply to eliminate fractions:

$$5(b-2)=3(r-4).$$ Expanding and rearranging terms gives:

$$5b-10=3r-12 \implies 3r-5b=2.$$ Now we solve the Diophantine equation $3r-5b=2$ for integers $r,b\geq7$.

From the equation $3r - 5b = 2$, by guess and check we find the smallest integer solution is $(r,b) = (4,2)$. We then get all solutions by increasing $r$ by $5$ and $b$ by $3$ each time: \begin{align*} r &= 4 + 5k, \\ b &= 2 + 3k. \end{align*}

With $r\ge7$, $b\ge7$, the five smallest values of $n$ are \begin{align*} k &= 2: & n &= 22, \\ k &= 3: & n &= 30, \\ k &= 4: & n &= 38, \\ k &= 5: & n &= 46, \\ k &= 6: & n &= 54. \end{align*}

Finally we sum the five values:

$$22+30+38+46+54=\boxed{190}.$$ ~Steven Zheng

Solution 2

Assuming there are $a$ red marbles, we write the original expression for the probability as

$$\dfrac{C_a^4\cdot C_{n-a}^3}{C_n^7}=\dfrac{C_a^5\cdot C_{n-a}^2}{C_n^7}.$$ Canceling the denominators and rewriting in factorial form gives

$$\dfrac{a!}{4!\cdot(a-4)!}\cdot\dfrac{(n-a)!}{3!\cdot(n-a-3)!}=\dfrac{a!}{5!\cdot(a-5)!}\cdot\dfrac{(n-a)!}{2!\cdot(n-a-2)!}.$$ Simplifying yields $3(a-4)=5(n-a-2)$, i.e., $8a-2=5n$, with $a,n-a\ge7$. The smallest solution is easily found to be $a,n=14,22$, and the smallest $5$ values of $n$ are $22,30,38,46,54$, whose sum is the final answer $\fbox{190}$. ~Confringo

Solution 3 (easy to understand but a little bashy)

Let there be $r$ red marbles and $b$ blue marbles. Consider the probability of getting $4$ red marbles and $3$ blue marbles. This is

$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} \cdot \frac{r-2}{r+b-2} \cdot \frac{r-3}{r+b-3} \cdot \frac{b}{r+b-4} \cdot \frac{b-1}{r+b-5} \cdot \frac{b-2}{r+b-6}$$ but we multiply by $\binom{7}{4}$ for the possible arrangements. Now, we do something clever when calculating the probability of getting $5$ red marbles and $2$ blue marbles. This probability is

$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} \cdot \frac{r-2}{r+b-2} \cdot \frac{r-3}{r+b-3} \cdot \frac{b}{r+b-4} \cdot \frac{b-1}{r+b-5} \cdot \frac{r-4}{r+b-6}$$ The clever thing was picking the marbles in the other red-red-red-red-blue-blue-red so it is easier to see that things cancel. We of course multiply by $\binom{7}{5}$. Thus, upon cancellation, we obtain

$$35 \cdot \frac{b-2}{r+b-6} = 21 \cdot \frac{r-4}{r+b-6} \rightarrow 5(b-2)=3(r-4) \rightarrow 3r-5b=2$$ Now, we can use moduli to determine that $3r-5b \equiv -5b \equiv b \pmod{3}$, so $b \equiv 2 \pmod{3}$. This means that the smallest solution is $(r, b)=(4, 2)$. Now, if $(r, b)$ is a solution, then $(r+5, b+3)$ is also a solution because $3(r+5)-5(b+3)=3r-5b$. Thus, we can generate solutions, where the five smallest ones with $b \ge 7$ and $r \ge 7$ are $(14, 8)$, $(19, 11)$ $(24, 14)$, $(29, 17)$. and $(34, 20)$. This means the sum of the five smallest values of $n$ are $22+30+38+46+54=\fbox{190}$