# 8.4. (7 points)

We will call a bus ticket with a six-digit number lucky if the sum of the digits of its number is divisible by 7. Can two consecutive tickets be lucky? If such tickets exist, provide examples of the numbers of such tickets.
Answer: for example, 429999 and 430000.

Solution: If two consecutive numbers differ only in the last digit, then their sums of digits differ by 1, and therefore cannot both be divisible by 7 simultaneously. Another case is when adding 1 to a number turns several nines at the end into zeros, and the next digit increases by 1 (for example, 699 and 700). How many nines do we need? If there is one nine, the sums of digits differ by $9-1=8$, if there are two, then by $2 \cdot 9-1=17$, if three, then by $3 \cdot 9-1=26$, if four, then by $4 \cdot 9 -1=35$, if five, then by $5 \cdot 9-1=44$, and so on. We need this difference to be divisible by 7. This is satisfied, for example, with four zeros at the end of the larger number. It remains to come up with a "beginning" for it: 429999 and 430000.

Comment. A correct example without justification - 3 points.