Czech and Slovak Mathematical Olympiad

Let $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$ be the prime factorization of a satisfactory number $n$. Here $p_1 < p_2 < \dots < p_k$ are all the prime divisors of $n$ and the exponents $\alpha_i$ are positive integers. The given equation implies that $p_1 = 2$ (otherwise $D = n$ which contradicts to $n = 3D + 5d$) and that $k \ge 2$ (otherwise $n$ is a power of $2$). Thus we have $D = p_2^{\alpha_2} \dots p_k^{\alpha_k}$, $d = p_2$ and the equation becomes
$$2^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k} = 3 p_2^{\alpha_2} \dots p_k^{\alpha_k} + 5p_2 \quad \text{or} \quad (2^{\alpha_1} - 3) p_2^{\alpha_2 - 1} \dots p_k^{\alpha_k} = 5.$$
(In the case when $k=2$ the left-hand of the last equation is simply $(2^{\alpha_1} - 3)p_2^{\alpha_2-1}$.)
Since the number $5$ has only two divisors $1$ and $5$, it holds that $2^{\alpha_1} - 3 \in \{1, 5\}$ and
hence either $\alpha_1 = 2$ or $\alpha_1 = 3$.

i. The case $\alpha_1 = 2$. The simplified equation
$$p_2^{\alpha_2-1} \dots p_k^{\alpha_k} = 5$$
holds if and only if either $k=2$, $p_2=5$ and $\alpha_2 - 1 = 1$, or $k=3$, $\alpha_2 - 1 = 0$, $p_3 = 5$ and $\alpha_3 = 1$ — then from $2 < p_2 < p_3 = 5$ it follows that $p_2 = 3$. Consequently, there are exactly two solutions in the case (i), namely $n = 2^2 5^2 = 100$ and $n = 2^2 3^1 5^1 = 60$.

ii. The case $\alpha_1 = 3$. The simplified equation
$$p_2^{\alpha_2-1} \dots p_k^{\alpha_k} = 1$$
holds only for $k=2$ and $\alpha_2 - 1 = 0$. Notice that there is no restriction on the prime number $p_2$ excepting the inequality $p_2 > 2$. Consequently, there are infinitely many solutions in the case (ii) and all of them are given by $n = 2^3 p_2^1 = 8p_2$, where $p_2$ is any odd prime number.

Answer. All the solutions $n$ are: $n = 60$, $n = 100$ and $n = 8p$, where $p$ is any odd prime number.

The given equation $n = 3D + 5d$ implies that $n > 3D$ and $n \le 3D + 5D = 8D$ (because of $d \le D$). Since the ratio $n:D$ must be a power of $2$, it follows from $3 < n: D \le 8$ that either (i) $n = 4D$, or (ii) $n = 8D$.

i. The case $n = 4D$. From $4D = n = 3D + 5d$ we have $D = 5d$ and thus $n = 4D = 20d$. Since $d$ must be a prime odd divisor of $n$ which is a multiple of $5$, we conclude that $p \in \{3, 5\}$ and hence $n \in \{60, 100\}$ (both the values are clearly satisfactory).

ii. $n = 8D$. Our way of deriving the inequality $n \le 8D$ implies now that $D = d$ and hence $D$ is an (odd) prime number. All such $n = 8D$ are solutions indeed.