## 1. Six-digit number

Each digit of the six-digit number, starting from the thousands place, is equal to the sum of the two preceding digits (which are to its left). What is the three-digit ending of the largest number with this property?

## Result: $\quad 369$
Solution, first method.

If the leading digit is 1, we have the following possibilities: 101123 and 112358.

Indeed, if the second digit from the left were 2, then the first five digits from the left would be $1,2,3,5,8$, and we could no longer choose a digit of one to satisfy the conditions of the problem. In particular, it is not possible for the second digit from the left to be greater than 2.

If the leading digit is 2, we have the possibility 202246.

Indeed, if the second digit from the left were 1, then the first five digits from the left would be $2,1,3,4,7$, and we could no longer choose a digit of one to satisfy the conditions of the problem. In particular, it is not possible for the second digit from the left to be greater than 1.

Similarly, we conclude that the only possibility for the leading digit to be 3 and the second from the left to be 0; thus we get the number 303369.

Similarly, we conclude that the leading digit cannot be 4 or greater than 4.

We see that the largest such number is 303369, and the last three digits are 369.

Solution, second method.

Let $\overline{a b c d e f}$ be a six-digit number with the given property.

Then for the digits $c, d, e$, and $f$, we have: $c=a+b, d=b+c, e=c+d, f=d+e$.

From all these equalities, it follows that $f=5 b+3 a$.

From the condition $3 a+5 b \leq 9$, it follows that $b=1$ or $b=0$. Meanwhile, $a$ is larger (and thus the six-digit number is larger) if $b=0$. Then the digit $a$ can be 1, 2, or 3. The largest six-digit number is obtained for $a=3$. If $a=3$, the other digits are: $c=3, d=3, e=6$, and $f=9$. This is the number 303369, so the desired three-digit ending is 369.