# Task 3. (12 points)

Egorov decided to open a savings account to buy a car worth 900000 rubles. The initial deposit amount is 300000 rubles. After one month and subsequently every month, Egorov plans to top up his account by 15000 rubles. The bank accrues interest monthly at an annual rate of $12 \%$. Interest accrued for the month is credited to the account, and in the following month, interest is also accrued on this interest. What is the minimum number of months required for the account to have a sufficient amount to buy the car?
# Solution

Let $S_{n}$ be the sum of the deposit after $n$ months, after interest accrual and after making additional contributions $D (15000$ rubles).

Since the bank accrues $1 \%$ per month, then

$$
\begin{gathered}
S_{1}=300000(1+0.01)+D \\
S_{2}=S_{1}(1+0.01)+D=\left(S_{0}(1+0.01)+D\right)(1+0.01)+D=S_{0}(1+0.01)^{2}+D(1+(1+0.01)) \\
S_{3}=S_{2}(1+0.01)=\left(S_{0}(1+0.01)^{2}+D(1+0.01)+D\right)(1+0.01)+D= \\
=S_{0}(1+0.01)^{3}+D\left((1+0.01)^{2}+(1+0.01)+1\right) \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
S_{n}=S_{0}(1+0.01)^{n}+D\left((1+0.01)^{n-1}+(1+0.01)^{n-2}+\ldots+(1+0.01)+1\right)
\end{gathered}
$$

By the formula for the sum of $n$ terms of a geometric progression, we get

$$
(1+0.01)^{n-1}+(1+0.01)^{n-2}+\ldots+(1+0.01)+1=\frac{1.01^{n}-1}{1.01-1}=\frac{1.01^{n}-1}{0.01} .
$$

Therefore, $S_{n}=S_{0}(1+0.01)^{n}+D \frac{1.01^{n}-1}{0.01}$.

The required number of months satisfies the inequality

$$
\begin{gathered}
300000 \cdot 1.01^{n}+15000 \frac{1.01^{n}-1}{0.01} \geq 900000 \Leftrightarrow \\
\Leftrightarrow 3 \cdot 1.01^{n}+15\left(1.01^{n}-1\right) \geq 9 \Leftrightarrow \\
\Leftrightarrow 18 \cdot 1.01^{n} \geq 24 \Leftrightarrow 1.01^{n} \geq \frac{24}{18} \Leftrightarrow n \geq 28.91
\end{gathered}
$$

Thus, the sufficient amount for buying the car will be in the deposit after 29 months.

## Answer: 29.