Task 1 - 310921
a) Give a natural number \(n\) for which (in the decimal positional system) the condition is satisfied that both the cross sum of \(n\) and the cross sum of \(n+1\) are divisible by 10!

Verify that the number you have given satisfies this condition!

b) Give the smallest natural number that satisfies the condition mentioned in a)!

Prove for the number you have given that it is the smallest number with this condition!
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If the number \(n\) does not end in the digit Nine, then the cross sum of \(n+1\) is exactly one greater than that of \(n\), since only the last digit increases by one and all others remain the same. Therefore, both cross sums cannot be divisible by 10. If, on the other hand, \(n\) ends in exactly \(k\) Nines, then these Nines become Zeros when transitioning to \(n+1\), while the digit before these Nines (possibly a leading Zero) increases by one. The cross sum of \(n+1\) is thus obtained from that of \(n\) by subtracting \(k \cdot 9 - 1\).

For both cross sums to be divisible by 10, the cross sum of \(n\) and \(k \cdot 9 - 1\) must both be divisible by 10. The smallest \(k\) that satisfies the latter condition is \(k=9\). Therefore, \(n\) must end in at least nine Nines. This means \(n\) has a minimum cross sum of 81, and since it should have a cross sum divisible by 10, it must be at least 90.

The smallest number that ends in at least nine Nines and has a cross sum of at least 90 is 9,999,999,999. However, this number ends in exactly ten Nines, so the cross sum of its successor reduces to 1, which is not divisible by 10.

The next larger, and thus second smallest, number that ends in at least nine Nines and has a cross sum of at least 90 is \(n=18,999,999,999\). Indeed, the cross sum of \(n\) is \(1+8+9 \cdot 9=90\) and that of \(n+1=19,000,000,000\) is \(1+9+9 \cdot 0=10\), so this \(n\) satisfies the condition and is the smallest such number.