
1. Find all positive integers $n$ which are not powers of 2 and which satisfy the equation $n=3 D+5 d$, where $D$ (and d) denote the greatest (and the least) numbers among the all odd divisors of $n$ which are larger than 1.

(Tomáš Jurík)


Solution. Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$ be the prime factorization of a satisfactory number $n$. Here $p_{1}2$. Consequently, there are infinitely many solutions in the case (ii) and all of them are given by $n=2^{3} p_{2}^{1}=8 p_{2}$, where $p_{2}$ is any odd prime number.

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

Remark. Let us show that the above solution can be presented more simply, if the "full" prime factorization of $n$ is replaced by a "partial" factorization $n=2^{\alpha} p l$, in which $2^{\alpha}$ is the greatest power of 2 dividing $n, p$ is the smallest prime divisor of $n$ and and $l$ is and odd number which has no prime divisor smaller than $p$ (thus we
have either $l=1$, or $l \geqslant p$ ). Using this factorization, we can write $D=p l, d=p$ and hence our task is to solve the equation

$$
n=2^{\alpha} p l=3 p l+5 p \quad \text { or } \quad\left(2^{\alpha}-3\right) l=5 \text {. }
$$

It follows that either $l=1$ and $2^{\alpha}-3=5$, or $l=5$ and $2^{\alpha}-3=1$. In the first case we have $\alpha=3$ and thus $n=8 p$, where $p$ is any prime number. In the second case it holds that $l=5, \alpha=2$ and thus $n=20 p$, but $5=l \geqslant p$ implies that $p \in\{3,5\}$, so there are only two solutions $n \in\{60,100\}$.

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

(i) The case $n=4 D$. From $4 D=n=3 D+5 d$ we have $D=5 d$ and thus $n=4 D=20 d$. 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) $n=8 D$. Our way of deriving the inequality $n \leqslant 8 D$ implies now that $D=d$ a hence $D$ is an (odd) prime number. All such $n=8 D$ are solutions indeed.
