A 16-member geometric sequence consists of positive integers. The first 5 terms are 9-digit numbers, the next 5 terms are 10-digit numbers, followed by 4 terms that are 11-digit numbers, and finally 2 terms that are 12-digit numbers. Provide the sequence!
Let the $i$-th term of the sequence be $a_{i}$, and its ratio be $q$. The terms from $a_{1}$ to $a_{5}$ are 9-digit numbers, from $a_{6}$ to $a_{10}$ are 10-digit numbers, and from $a_{11}$ to $a_{14}$ are 11-digit numbers. Thus, the number of digits in $a_{10}$ is only 1 more than the number of digits in $a_{1}$, so $a_{10} = a_{1} q^{9} \cdot 10 \cdot a_{10}$, which means $a_{1} q^{14} > 10 \cdot a_{1} q^{9}$, hence $\lg q > 1 / 5 = 0.2$, and $q > 1.58$; summarizing, $1.581$, because $q$ cannot be an integer. Since $a_{16} = a_{1} q^{15} = a_{1} p^{15} / r^{15}$ is an integer, $r^{15}$ must be a divisor of $a_{1}$, so $a_{1} = k \cdot r^{15}$, where $k$ is a natural number, thus $r^{15} \leq a_{1} \cdot 1.67$. For the remaining case $r=3$, $p=5$ is the only suitable value because $4 / 3 \approx 2.38555$, so $\lg k \geq 8 - 7.15665 = 0.84335$, and $k > 6$. From the second condition,

$$
k < \frac{10^{10}}{3^{15} (5 / 3)^{9}} = \frac{2^{10} \cdot 5^{10}}{3^{6} \cdot 5^{9}} = \frac{2^{9} \cdot 10}{3^{6}} = \frac{5120}{729} < 8
$$

Therefore, $k=7$, so $a_{1} = 7 \cdot 3^{15} = 100442349$, and $a_{i} = 7 \cdot 3^{16-i} \cdot 5^{i-1}$. The last term, $a_{16} = 213623046875$, is no longer divisible by 3.

Iváncsy Szabolcs (Miskolc, Electrical Engineering, 4th year)