9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the notation of any number between them, there is an odd digit. Find the largest possible value of the difference $b-a$.

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Translation:
9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the notation of any number between them, there is an odd digit. Find the largest possible value of the difference $b-a$.
Answer: 111112.

Solution. Estimation. We will prove that to a 9-digit number $a$, less than 888 888, all digits of which are even, one can add a number not exceeding 111112 so that all its digits will again be even. If among the digits of the number $a$, except for the first one, there is a digit less than 8, then it can be increased by 2. Otherwise, the number has the form $A 88888$. Since $A<8$, one can add 111112 to it, and the result will be $(A+2) 00000$. If, however, $a=888$ 888, then all larger 6-digit numbers contain an odd digit, so there cannot be a suitable number $b$ among them.

Example. It remains to check that the difference 111112 is possible. The reasoning from the estimation suggests that examples could be the numbers $a=288888$ and $b=400$ 000. Indeed: any number between them has either the digit 3 or the digit 9.