63rd 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 $p_1<p_2<\cdots <p_k$ 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 $$2^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_k^{{\alpha }_k}=3p_2^{{\alpha }_2}\dots p_k^{{\alpha }_k}+5p_2\text{or}(2^{{\alpha }_1}-3)p_2^{{\alpha }_2-1}\dots p_k^{{\alpha }_k}=5.$$ (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.

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Alternative solution.

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.