Matěj had six different natural numbers written on a line in his notebook. The second of them was double the first, the third was double the second, and similarly, each subsequent number was double the previous one. Matěj copied all these numbers into the following table, in random order, one number per cell.

![](https://cdn.mathpix.com/cropped/2024_04_17_b3bac5a705eebcae18d0g-2.jpg?height=189&width=288&top_left_y=1756&top_left_x=884)

The sum of the two numbers in the first column of the table was 136, and the sum of the numbers in the second column was double that, i.e., 272. Determine the sum of the numbers in the third column of the table.

(L. Šimünek)
The smallest added number we denote by the unknown $a$ and use it to express all added numbers:

$$
a, 2 a, 4 a, 8 a, 16 a, 32 a \text {. }
$$

The number $32 a$ cannot be in the first column, because the sum of any other two added numbers is less than $32 a$, and thus the sum of the numbers in the second column could not be greater than in the first. Possible sums of the numbers in the first column are:

$$
\begin{array}{rlrlrl}
a+2 a & =3 a & 2 a+4 a=6 a & & 4 a+8 a=12 a & 8 a+16 a=24 a \\
a+4 a & =5 a & 2 a+8 a=10 a & 4 a+16 a=20 a & \\
a+8 a & =9 a & 2 a+16 a=18 a & & \\
a+16 a & =17 a & & &
\end{array}
$$

The natural number that multiplies the unknown $a$ in these sums must be a divisor of the number $136=$ $=2 \cdot 2 \cdot 2 \cdot 17$. This is satisfied only by the sum

$$
17 a=a+16 a .
$$

The sum of the numbers in the second column is then $34 a$ and this can be obtained only as

$$
34 a=2 a+32 a .
$$

For the third column, the remaining numbers are $4 a$ and $8 a$. From the equation $17 a=136$ we get $a=8$; the sum of the numbers in the third column is thus

$$
4 a+8 a=12 a=12 \cdot 8=96 \text {. }
$$

Evaluation. 3 points for the insight that in the first column are $a$ and $16 a$, including the justification that it is the only possibility; 1 point for the insight that in the third column are $4 a$ and $8 a$; 2 points for the result 96.