## Task 4 - 090914

The first cross sum of a natural number $z$ is understood to be the cross sum formed in the usual way. If the first cross sum of $z$ is a number with more than one digit, then its cross sum is called the second cross sum of $z$.

Examples: The first cross sum of 98 is $9+8=17$, the second cross sum of 98 is $1+7=8$. The first cross sum of 43 is $4+3=7$, a second cross sum of 43 is not defined.

If the second cross sum of $z$ is a number with more than one digit, then its cross sum is called the third cross sum of $z$. In a similar manner, higher cross sums are defined if applicable.

a) Determine the number of natural numbers from 10 to 1000 for which no second cross sum is defined!

b) Determine the number of natural numbers from 10 to 1000 for which the second, but not the third, cross sum is defined!

c) Determine the smallest natural number for which a fourth cross sum is defined!
a) For the 2nd cross sum not to be defined, the 1st cross sum must be less than 10. If we consider all numbers that can be formed from 3 digits (each digit from 0 to 9), this also includes all one- and two-digit numbers. The following digits lead to a cross sum of 0 to 9:

$\{0,0,0\}, \ldots\{0,0,9\} \Rightarrow 10$ possibilities $\{0,1,1\}, \ldots\{0,1,8\} \Rightarrow 8$ possibilities

$\{0,2,2\}, \ldots\{0,2,7\} \Rightarrow 6$ possibilities $\{0,3,3\}, \ldots\{0,3,6\} \Rightarrow 4$ possibilities

$\{0,4,4\}, \ldots\{0,4,5\} \Rightarrow 2$ possibilities $\{1,1,1\}, \ldots\{1,1,7\} \Rightarrow 7$ possibilities

$\{1,2,2\}, \ldots\{1,2,6\} \Rightarrow 5$ possibilities $\{1,3,3\}, \ldots\{1,3,5\} \Rightarrow 3$ possibilities

$\{1,4,4\} \Rightarrow 1$ possibility $\{2,2,2\}, \ldots\{2,2,5\} \Rightarrow 4$ possibilities

$\{2,3,3\}, \ldots\{2,3,4\} \Rightarrow 2$ possibilities $\{3,3,3\} \Rightarrow 1$ possibility

Of these 53 possibilities, there are 4 variants with 3 identical digits, 26 variants with one identical digit pair, and 23 variants with all different digits. The combination of different digits results in $6=3$!, so $23 \cdot 6=138$ different numbers; for a pair of identical digits, there are $3=3$! / 2!, so $26 \cdot 3=78$ numbers. Thus, in total (since 0 to 9 are not, but 1000 is also counted), the following number of sought numbers is obtained: $4+78+138-10+1=211$

b) 991 numbers are considered in total. 211 form only one cross sum. Therefore, there are 991 - $211=780$ numbers that form a second cross sum. Only those that form a third cross sum need to be excluded. The first cross sum can be at most $27=9+9+9$, meaning the 2nd cross sum is at most 10 (exactly when the 1st cross sum is 19). And only in these cases is the 3rd cross sum defined. We must therefore exclude all numbers whose 1st cross sum is 19.

No two-digit number has a cross sum of 19. The sought numbers are therefore all three-digit. The digits of the sought numbers consist of the following combinations and their permutations: $\{9,9,1\},\{9,8,2\},\{9,7,3\},\{9,6,4\},\{9,5,5\},\{8,8,3\},\{8,7,4\},\{8,6,5\},\{7,7,5\},\{7,6,6\}$. Among these 10 possibilities, there are 5 with different digits and 5 with one identical digit pair. Thus, analogous to a), the numbers that can be formed from these are: $5 \cdot 6+5 \cdot 3=30+15=45$ different numbers with a 3rd cross sum.

The sum of the numbers that form a 2nd but not a 3rd cross sum is therefore $780-45=735$

c) The smallest natural number for which a fourth cross sum is defined is:

19,999,999,999,999,999,999,999

There is no smaller one, as the cross sum must be kept as low as possible to keep the number as small as possible. The smallest 2nd cross sum can only be 19, as it is the smallest two-digit number that has another two-digit cross sum. Therefore, the 1st cross sum itself must have a cross sum of 19, and no smaller number than 199 can satisfy this. The 1st cross sum is formed by adding all the digits of the number. The smallest number that can form this cross sum has as few digits as possible, i.e., as many 9s as possible from the right. This leads to a number starting with 1 and followed by 22 nines.