- #26

- 63

- 13

From 1 to 31, all natural numbers can be written in one unique way in the following form $$ a\cdot 1 + b\cdot 2 + c\cdot 2^2 +d\cdot 2^3 + e \cdot 2^4 = a\cdot 1 +b\cdot 2 +c\cdot 4 +d\cdot 8+ e\cdot 16$$ wherein a, b, c, d, e are each 0 or 1.7.Five vessels contain ##100## balls each. Some vessels contain only balls of ##10\, g## mass, while the other vessels contain only balls of ##11\, g## mass. How can we determine with a single weighing which results in a mass, which vessels contain balls of ##10\, g## and which contain balls of ##11\, g##? (It is allowed to remove balls from the vessels.)

In the first vessel, we leave one weight, in the second vessel 2, 4 in the third, 8 in the fourth and 16 in the fifth. We weigh all of these weights. Since each weight is either 10 or 11 g, if we subtract 310 g, then we get exactly the weight

*that the 11 g weights contributed to the total weight, with their weight above 10 g*. The difference is expressed in the above form, and a, b, c, d, e show which vessels contributed to the total weight by weight over 10 g.

Let's look at an example: First vessel: 10 g; second 11 g; third 11 g; fourth 10 g; fifth 11g. Using the above method, the total weight ## 1 \cdot 10 + 2 \cdot 11 +4 \cdot 11 +8 \cdot 10 +16 \cdot 11 = 332##. Furthermore ##332-310 =22 =2+4+16.##