Karpenkov O.

To open the safe, you need to enter a code - a number consisting of seven digits: twos and threes. The safe will open if there are more twos than threes, and the code is divisible by both 3 and 4. Come up with a code that opens the safe.

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Use the divisibility rule for 3.

## Solution

Since there are more twos than threes, the number of twos can be 4, 5, 6, or 7. In the first case, the sum of the digits is 17, in the second case 16, in the third case 15, and in the last case 14. According to the divisibility rule for 3, only the third option works.

Thus, the code contains 6 twos and one three. According to the divisibility rule for 4, the number formed by the last two digits is 32.

## Answer

2222232.