## Task 4 - 280724

a) Determine the sum of the cross sums of all two-digit numbers divisible by 5!

b) Determine the sum of the cross sums of all natural numbers from 0 to 1000!
a) For each possible tens digit, there are exactly two numbers divisible by 5, namely one that ends in 0 and one that ends in 5.

The sum of the tens digits of all two-digit numbers divisible by 5 is thus \(2 \cdot (1 + 2 + 3 + \ldots + 9) = 90\), and the sum of their units digits is obtained with \(9 \cdot 5 + 9 \cdot 0 = 45\). Therefore, the desired sum of the mentioned digit sums is \(90 + 45 = 135\).

b) If we consider all natural numbers from 0 to 999, we recognize that each of the digits from 0 to 9 appears exactly 100 times in the hundreds place.

If we leave the hundreds place unchanged, each of the digits appears exactly ten times in the tens place, and when running through all ten digits of the hundreds place, this totals \(10 \cdot 10 = 100\) times. If we leave the tens place and the hundreds place unchanged, each of the digits appears exactly once in the units place, totaling \(1 \cdot 10 \cdot 10 = 100\) times. The number 1000 has the digit sum of 1.

From the aforementioned observations, it follows: The desired sum of the mentioned digit sums is

\[
300 \cdot 1 + 300 \cdot 2 + \ldots + 300 \cdot 9 + 1 = 300 \cdot 45 + 1 = 13501
\]

Solutions of the 2nd Round 1988 adopted from [5]

### 4.30.3 3rd Round 1988, Class 7