Would you like to take such a loan?

Let us think of one year as one unit of time.

Then the interest per unit time that we need to give is $0.1P(t)$ per unit time. This means that the change in the amount that we need to pay after infinitely small time $dt$ is given by:

$P(t+dt)-P(t)=0.1P(t)dt$

$\therefore \frac{P(t+dt)-P(t)}{dt}=0.1P(t)$

$\therefore \frac{dP}{dt}=0.1P(t)$

$\therefore P(t)=P(0)e^{(0.1)t}$

Hence for $t=40$ , we get,

$P(t)=54.5981P(0)$

This gives $\lfloor k \rfloor = \boxed{54}$

The formula for compound interest is

$A = P \left( 1 + \dfrac{r}{n} \right)^{rt}$ where

• $A$ is the final amount
• $P$ is the principal amount
• $r$ is the annual interest rate
• $t$ is the number of years for which the money is deposited
• $n$ is the amount of times the money is compounded annually.

Now, as the money is compounded in infinitely small units of time, $n \to 0$ . Thus, we need to find out the formula when $n \to 0$ . We shall use limits. The formula we obtain for continuous compounding is

$A = P e^{rt}$ Check this link on Khan academy for its derivation

Now, the final amount is $kP$ , the initial amount is $P$ and r is $10\%$ or $0.1$ and t is $40$ .

Plugging in the values,

$kP = P e^{40 \times 0.1}$

$kP = P e^{4}$

$k = e^{4}$

$k = 54.59$ , and thus $\lfloor k \rfloor = \boxed{54}$