131. A natural number $n$ is the product of four consecutive natural numbers, each greater than 5. Determine the maximum number of possible last digits of $n$, given that its last digit is not 0.
Solution. Among four consecutive numbers whose product is equal to $n$, there are two even numbers, and therefore the number $n$ is even. Consequently, none of these numbers is divisible by 5 - otherwise, the number $n$ would end in the digit 0. Thus, we have:

$$
\begin{gathered}
n=(5 k+1)(5 k+2)(5 k+3)(5 k+4)=\left(25 k^{2}+25 k+4\right) \times \\
\times\left(25 k^{2}+25 k+6\right)=(25 k(k+1)+4)(25 k(k+1)+6)= \\
=(50 m+4)(50 m+6)
\end{gathered}
$$

where $m=\frac{k(k+1)}{2}$ is an integer. From this, we get:

$$
\begin{gathered}
n=2500 m^{2}+500 m+24=2000 m^{2}+500 m(m+1)+24= \\
=1000\left(2 m^{2}+p\right)+24
\end{gathered}
$$

where $p=\frac{m(m+1)}{2}$ is an integer.

Thus, the last three digits of the number $n$ are determined - they are 024. The fourth digit from the end of the number is not uniquely determined; for example, $6 \cdot 7 \cdot 8 \cdot 9=3024,11 \cdot 12 \cdot 13 \cdot 14=24024$.