1.1. On September 1, 2021, Vasya deposited 100,000 rubles in the bank. Exactly one year later, the bank accrues 10% annual interest (that is, it increases the amount by 10 percent of what was on the account at that moment, for example, on September 2, 2022, Vasya will have 110,000 rubles on his account). Find the smallest year number in which on September 2, the amount on Vasya's account will be more than 150,100 rubles.
Answer: 2026

Solution. Note that on September 2nd of the $(2021+n)$-th year, the account will have $100000 \cdot 1.1^{n}$ rubles. Since $100000 \cdot 1.1^{4}=146410<150100<161051=100000 \cdot 1.1^{5}$, the minimum $n$ for which the amount will exceed 150,100 rubles is 5. Therefore, the answer is: $2021+5=2026$.