Problem 4.8. Leonid had 7 bags of sugar weighing $3,7,12,15,17,28,30$ kilograms. Over three days, he sold all the bags except one. It turned out that the weight of the sugar sold on the second day was twice that of the first day; and on the third day, it was twice that of the second day. What is the weight of the bag that remained unsold? List all possible options.
Answer: 7 and 28.

Solution. Let $x$ kilograms of sugar be sold by Leonid on the first day. Then on the second day, he sold $2x$, and on the third day, $4x$ kilograms of sugar. This means the weight of the sold sugar is $x + 2x + 4x = 7x$ kilograms, which is divisible by 7.

Notice that the total weight of the sugar, $3 + 7 + 12 + 15 + 17 + 28 + 30 = 112$ kilograms, is also divisible by 7. Therefore, the weight of the remaining bag is also divisible by 7. Hence, Leonid did not sell either 7 kilograms or 28 kilograms of sugar.

It remains to provide examples to prove that both options are possible.

- On the first day, Leonid sold $3 + 12 = 15$ kilograms, on the second day 30 kilograms, on the third day $15 + 17 + 28 = 60$ kilograms, and the bag with 7 kilograms remained.
- On the first day, Leonid sold 12 kilograms, on the second day $7 + 17 = 24$ kilograms, on the third day $3 + 15 + 30 = 48$ kilograms, and the bag with 28 kilograms remained.