7.3. Egor borrowed 28 rubles from Nikita, and then returned them in four payments. It turned out that Egor always returned a whole number of rubles, and the amount of each payment increased and was divisible by the previous one. What was the last amount Egor paid?
Answer: 18 rubles

Solution:

1) If Egor paid $a$ rubles the first time, then the second time - not less than $2a$, the third - not less than $4a$, the fourth - not less than $8a$, and in total - not less than $15a$. Since $15a \leq 28$, we get that $a=1$.
2) The second time he paid 2 or 3 rubles (because if 4, then he paid at least $1+4+8+16=29>28$).

2.1) If he paid 2 rubles, then he had 25 rubles left to pay, and in this case, all subsequent payments would be even. This case is impossible.

2.2) Therefore, he paid 3 rubles, and for the last two times, he had 24 rubles left to pay. Let's say he paid $a$ times more the third time than the second, and $b$ times more the fourth time than the third. Then $3a + 3ab = 24$, which means $a + ab = a(b+1) = 8$. It follows that $a$ and $b+1$ are powers of two, and $b>1$, which is only possible with $a=2$ and $b=3$. Hence the answer.

## Criteria:

Only the answer $0-0$ points.

Only the answer with verification (all 4 sums are written out, showing that all conditions are met) - 1 point, it does not add up with anything.

Each of the points 1), 2), 2.1) is worth one point, these points are added up. A generally correct solution, but 2.2) is proven by incomplete enumeration - no more than 5 points.