Question

По пути домой Коля и Таня предложили родителям отгадать номер их входных билетов в музей. Чтобы было легче отгадывать, Коля сказал, что:- номера билетов является четырёхзначными числами, составленными из цифр2 6 7- ёщё одна цифра либо 0, либо 5;- цифра встречается в номере билета один раз;- первая цифра числа равна 6;- это число делится на 3;- это число делится на 4.Запишите номера входных билетов ребят.

Answer 1

Число делиться на 4 если последние 2 цифры деляться на 4 
число делиться на 3 если сумма всех цифр делиться на 3 , а значить что в этом числе не может быть 5 а значить их номера : 
6072 и 6720

Answer 2

Problem Analysis

Kolya and Tanya are trying to guess the number of their entrance tickets to the museum. To make it easier, Kolya provides the following clues about the number: - The number is a four-digit number composed of the digits 2, 6, and 7, along with one additional digit, which can be either 0 or 5. - The additional digit appears only once in the number. - The first digit of the number is 6. - The number is divisible by 3. - The number is divisible by 4.

Solution

Let's analyze the given clues to find the possible numbers for their entrance tickets.

1. The number is a four-digit number composed of the digits 2, 6, and 7, along with one additional digit, which can be either 0 or 5. - This means the number can be in the form of 6XYZ, where X can be either 2, 6, or 7, and Y and Z can be either 0 or 5.

2. The additional digit appears only once in the number. - This means that X, Y, and Z cannot be the same.

3. The first digit of the number is 6. - This means X must be 2 or 7.

4. The number is divisible by 3. - For a number to be divisible by 3, the sum of its digits must be divisible by 3. - Since the number is in the form 6XYZ, the sum of its digits is 6 + X + Y + Z. - The sum of the digits must be divisible by 3, so (6 + X + Y + Z) % 3 = 0.

5. The number is divisible by 4. - For a number to be divisible by 4, the last two digits must form a number divisible by 4. - Since the number is in the form 6XYZ, the last two digits are YZ. - The number YZ must be divisible by 4, so (10 * Y + Z) % 4 = 0.

Let's go through the possible combinations of X, Y, and Z to find the valid numbers.

Possible Numbers

1. X = 2, Y = 0, Z = 5 - The number is 6205. - The sum of the digits is 6 + 2 + 0 + 5 = 13, which is not divisible by 3. - The number is not valid.

2. X = 2, Y = 5, Z = 0 - The number is 6250. - The sum of the digits is 6 + 2 + 5 + 0 = 13, which is not divisible by 3. - The number is not valid.

3. X = 2, Y = 5, Z = 5 - The number is 6255. - The sum of the digits is 6 + 2 + 5 + 5 = 18, which is divisible by 3. - The number is valid. - The last two digits, 55, form a number divisible by 4. - The number is valid.

4. X = 7, Y = 0, Z = 5 - The number is 6705. - The sum of the digits is 6 + 7 + 0 + 5 = 18, which is divisible by 3. - The number is valid. - The last two digits, 05, form a number divisible by 4. - The number is valid.

5. X = 7, Y = 5, Z = 0 - The number is 6750. - The sum of the digits is 6 + 7 + 5 + 0 = 18, which is divisible by 3. - The number is valid. - The last two digits, 50, form a number divisible by 4. - The number is valid.

6. X = 7, Y = 5, Z = 5 - The number is 6755. - The sum of the digits is 6 + 7 + 5 + 5 = 23, which is not divisible by 3. - The number is not valid.

Conclusion

Based on the given clues, the valid numbers for Kolya and Tanya's entrance tickets are 6255, 6705, and 6750.

Please note that these are the possible valid numbers based on the given clues. There may be other valid numbers that satisfy the conditions as well.