## Task 3 - 120823

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

a) Determine the largest value that can occur as the third cross sum of a 1972-digit number!

b) Give (by describing the digit sequence) the smallest 1972-digit natural number that has this largest possible third cross sum!
a) Among all 1972-digit numbers, the one with the largest first cross sum consists of 1972 digits of 9. This largest first cross sum is thus $9 \cdot 1972 = 17748$.

Since the first cross sum of a 1972-digit number cannot be greater than 17748, the largest second cross sum is obtained as the largest of all cross sums of the numbers from 10 to 17748, and consequently as the cross sum of the number 9999, i.e., 36.

Since the second cross sum of a 1972-digit number cannot be greater than 36, the largest third cross sum is obtained as the largest of all cross sums of the numbers from 1 to 36, and consequently as the cross sum of the number 29, i.e., 11.

b) Among all numbers from 10 to 36, 29 is the only one with the cross sum 11. Among all numbers with the cross sum 29, one finds the smallest by forming a number with the last three digits 9 such that the number formed by the preceding digits is the smallest with the cross sum 2. This number is clearly 2999.

With a similar justification, one finds the smallest 1972-digit number $A$ with the cross sum 2999, because $\frac{2999}{9} = 333 \frac{2}{9}$, by forming a number with the last 333 digits 9 such that the number formed by the preceding 1972 - 333 = 1639 digits is the smallest 1639-digit number with the cross sum 2. Thus, the number

$$
A = 1 \underbrace{0 \ldots 0}_{1637} 1 \underbrace{9 \ldots 9}_{1637}
$$

consists in sequence of one digit 1, 1637 digits 0, one digit 1, and 333 digits 9.

If $g$ is any number different from the smallest number 2999, i.e., a larger number with the cross sum 29, and if a 1972-digit number $n$ has the number $g$ as its cross sum, then:

Since $\frac{g}{9} \geq \frac{3000}{9} = 333 \frac{3}{9}$, the smallest 1972-digit number $B$ with the cross sum $g$ has as its last 333 digits each a 9 and among the preceding 1639 digits at least one digit $\geq 2$. Therefore, $B > A$ and thus certainly $n > A$.

Thus, $A$ is proven to be the smallest among all 1972-digit numbers that have any number with the cross sum 29 as their cross sum, i.e., but also, that have the third cross sum 11.