Stars on the picture represent 16 consecutive natural multiples of the number three. The numbers in the boxes have the same sum. Determine the smallest of these 16 numbers.

(L. Šimünek)

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Let's compare the last, i.e., the largest numbers of both frames: the largest number of the second frame lies 10 places to the right of the largest number of the first frame, so it is 30 greater. By the same comparison, we find that the second largest numbers in the frames also differ by 30, as do the third largest, the fourth largest, and the fifth largest numbers. The sum of all five numbers in the second frame is therefore $5 \cdot 30=150$ greater than the sum of the five largest numbers in the first frame. For the sums in the frames to be equal, the remaining number in the first frame must be exactly 150. The smallest number in this sequence is therefore 150.

Another solution. First, let's find 16 consecutive natural numbers that satisfy the condition of equal sums. If we multiply the found numbers by three, the sums in the frames will also be multiplied by three, and their equality will be preserved. By this method, we will reach the desired result and simplify the calculations, as we will initially work with smaller numbers.

Let the smallest number in the sequence of consecutive natural numbers be $x$ and express the sum of the numbers in the first frame:

$$
x+x+1+x+2+x+3+x+4+x+5=6 x+15 .
$$

The first number in the second frame is $x+11$ in the given notation. Express the sum of the numbers in the second frame:

$$
x+11+x+12+x+13+x+14+x+15=5 x+65 .
$$

By solving the equation $6 x+15=5 x+65$, we get $x=50$. By multiplying by three, we get the first number in the sequence of multiples of three, which is $50 \cdot 3=150$.

Evaluation. 2 points for the result; 4 points for explaining the method.