3. What digits does the binomial coefficient $\binom{n}{4}, n=4,5,6, \ldots$ end with?
Solution. We have

$$
\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24}
$$

Every natural number can be represented in one of the forms

$$
5 k, \quad 5 k+1, \quad 5 k+2, \quad 5 k+3, \quad 5 k+4
$$

If the number $n$ is of the form $5 k$ or $5 k+1$ or $5 k+2$ or $5 k+3$, then the number $n(n-1)(n-2)(n-3)$ is divisible by 5. Since 5 and 24 are coprime, it follows that $\binom{n}{4}$ is divisible by 5, so $\binom{n}{4}$ ends in 0 or 5.

If the number $n$ is of the form $5 k+4$, then substituting in (1), we will get that

$$
\binom{n}{4}=\frac{5 m}{24}+1
$$

using the fact that 5 and 24 are coprime and that $\frac{5 m}{24}$ is a natural number, it follows that $\frac{5 m}{24}$ is divisible by 5, i.e., it ends with 0 or 5. According to this, the number $\binom{n}{4}, n \geq 5$ will end in 1 or 6.