Ответ:

При делении на число m возможны остатки от 0 до m − 1. Выпишем числа, начиная с наименьших натуральных чисел в каждом классе.

Деление на 3:

  1. Остаток 0: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
  2. Остаток 1: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28.
  3. Остаток 2: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29.

Деление на 4:

  1. Остаток 0: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40.
  2. Остаток 1: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37.
  3. Остаток 2: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38.
  4. Остаток 3: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39.

Деление на 7:

  1. Остаток 0: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70.
  2. Остаток 1: 1, 8, 15, 22, 29, 36, 43, 50, 57, 64.
  3. Остаток 2: 2, 9, 16, 23, 30, 37, 44, 51, 58, 65.
  4. Остаток 3: 3, 10, 17, 24, 31, 38, 45, 52, 59, 66.
  5. Остаток 4: 4, 11, 18, 25, 32, 39, 46, 53, 60, 67.
  6. Остаток 5: 5, 12, 19, 26, 33, 40, 47, 54, 61, 68.
  7. Остаток 6: 6, 13, 20, 27, 34, 41, 48, 55, 62, 69.