# Task 4.

Find the smallest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal representation.
Answer: 100023457896

Solution. The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. Since we need the smallest number, we will add two digits 0 and place the number 10002345 at the beginning of the required number (this is the smallest possible "start" of the number). The remaining digits $6,7,8$ and 9 need to be arranged so that the number is divisible by 4. This means the last two digits should be 68, 76, or 96. The minimum is 7896.

## B-2

Find the smallest 13-digit natural number that is divisible by 75 and contains all 10 digits in its notation.

Answer: 1000023468975

Solution. The number is divisible by 3 and 25. The sum of ten digits is 45 (divisible by three), so the sum of the three additional digits in the number's notation should be $0,3,6, \ldots, 27$. The older the digit place, the smaller the digit it is better to put in it, so we will take three zeros and place the number 10000 at the beginning of the required number. To make the number divisible by 25, the last two digits should be 25, 50, or 75. The first two options use small digits, which are better placed in older places, so we choose 75 and get the answer: 1000023468975.

## B-3

Find the largest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal notation.

Answer: 999876543120

The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. The older the digit place, the larger the digit it is better to put in it, so we will add two nines to the ten digits and start the number as: 999.... For divisibility by 4, the last two digits should form a number divisible by 4: $00,04,08,12,16,20,24, \ldots, 96$. Of these options, 20 is the best because it leaves the larger digits for the older places (we have only one zero, so 00 is out). In the end, we get the number 999876543120.

## B-4

Find the largest 13-digit natural number that is divisible by 75 and contains all 10 digits in its notation.

Answer: 9999876432150

Solution. The number is divisible by 3 and 25. The sum of ten digits is 45 (divisible by three), so the sum of the three additional digits in the number's notation should be $0,3,6, \ldots, 27$. The older the digit place, the larger the digit it is better to put in it, so we will take three nines and start the number as: $9999 \ldots$. To make the number divisible by 25, the last two digits should be 25, 50, or 75. The five is used in any case, and it is best to choose 50 to leave the larger digits 7 and 2 for the older places. In the end, we get the number 9999876432150.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary Stage 2020/21 academic year for 5-6 grades

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