## Task 6 - 060836

Consider the product of all odd numbers that are greater than 30 and less than 50. Without fully calculating the product, answer the following questions:

a) What digit is in the units place of the product?

b) Is the product an 18-digit number?
a) If you multiply a number that ends in 5 by any odd number, the result will again be a number with the units digit 5. The product to be investigated contains the factor 45, which ends in 5, and only odd factors otherwise. Therefore, its units digit is 5.

b) The product can be thought of as being split into the partial products $31 \cdot 49, 33 \cdot 47, 35 \cdot 45, 37 \cdot 43$, and $39 \cdot 41$. Each of these partial products is of the form $(a-b)(a+b)$ with $a=40$ and $1 \leq b \leq 9$, and is therefore a positive number that is certainly less than 2000. The entire product is thus less than $2000^{5}$.

However,

\[
\begin{aligned}
2000^{5} & =2000 \cdot 2000 \cdot 2000 \cdot 2000 \cdot 2000 \\
& =2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 1000 \cdot 1000 \cdot 1000 \cdot 1000 \cdot 1000 \\
& =32 \cdot 1000000000000000
\end{aligned}
\]

is a 17-digit number. Therefore, the product to be investigated cannot be an 18-digit number.

Solutions of the III. Round 1966 taken from [5]

### 5.9 VII. Olympiad 1967

### 5.9.1 I. Round 1967, Class 8