Алгоритмы с обратным преобразованием
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- Ruby
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Description
def find_numbers_by_digit(digit, max_val=100):
return [10 * m + digit for m in range((max_val // 10) + 1) if 10 * m + digit <= max_val]
# find_numbers_by_digit(3, 50) → [3, 13, 23, 33, 43]
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