Razin M.

There is a set of 20 weights with which any integer weight from 1 to 1997 g can be measured (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the minimum possible weight of the heaviest weight in such a set, if:

a) the weights in the set are all integers,

b) the weights do not necessarily have to be integers

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Let's order the weights of the weights (in grams) in ascending order: $p_{0} \leq p_{1} \leq \ldots \leq p_{19}$. Clearly, $p_{0} \leq 1$ (otherwise, it's impossible to weigh a load of 1 g). Similarly, $p_{1} \leq 2$ (to weigh 2 g). Since these two weights are not enough to weigh 4 g, $p_{2} \leq 4=2^{2}$. Continuing this reasoning, we arrive at the inequality $p^{7} \leq 128=2^{7}$.

a) The weight of the eight lightest weights does not exceed $1+2+2^{2}+\ldots+2^{7}=255$. The weight of the remaining 12 (heavier) weights is at least $1997-255=1742$, and thus, the weight of the heaviest one is at least $1742: 12 > 145$.

On the other hand, a set of weights of 1, 2, 4, ..., 128 g, ten weights of 145 g each, and two weights of 146 g each satisfies the conditions of the problem, as with weights of 1, 2, 4, ..., 128 g, any weight from 1 to 255 g can be formed (see problem 30839).

b) Suppose the heaviest weight weighs less than 145.25 g. Then the total weight of the 12 heaviest weights is less than $145.25 \cdot 12 = 1743$. Since

$p_{0}+\ldots+p_{6} \leq 127$, then $p_{7} > 1997 - 1743 - 127 = 127$. Therefore, $127 < p_{7} \leq 128$. At the same time, $p_{8} > 1997 - 255 - 11 \cdot 146 > 128$. Therefore, $p_{7} = 128$ (otherwise, it's impossible to weigh 128 g), which means all the light weights have integer weights, their total weight is 255, and the weight of all the heavy weights is at least $1997 - 255 = 1742$. Let's try to weigh a load of 700 g. We will have to use exactly four heavy weights. Indeed, three are insufficient ($255 + 3 \cdot 146 < 700$), and five are too many: any five heavy weights weigh more than $7 \cdot 146 = 720$. The total weight of these four weights is an integer. If it is not less than 581, then one of these four weights weighs at least $581: 4 = 145.25$. If it is not more than 580, then the weight of one of the remaining eight heavy weights is at least

$(1742 - 580): 8 = 145.25$. In any case, the weight of the heaviest weight is not less than 145.25 g. Contradiction.

On the other hand, a set consisting of weights of 1, 2, 4, ..., 128 g, four weights of 145 g each, and eight weights of 145.25 g each suits us. Weights of 145.25 g are used only in sets of four (the weight of such a set is 581 g). Any integer load up to 255 g can be weighed using the eight light weights. If the weight of the load is within the range from 1 to $835 = 255 + 4 \cdot 145$, the necessary number of weights of 145 g is used. By adding one or two sets, any weight from 1 to $2 \cdot 581 + 835 = 1997$ g can be obtained.

## Answer

a) 146 g;
b) 145.25 g.

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