def lcm(x, y):
    """
    计算两个数的最小公倍数
    """
    return abs(x*y) // math.gcd(x, y)


def lcm_of_three(x, y, z):
    """
    计算三个数的最小公倍数
    """
    # 首先计算a和b的最小公倍数
    lcm_xy = lcm(x, y)
    # 然后将结果与c计算最小公倍数
    return lcm(lcm_xy, z)


if __name__ == "__main__":
    solutions = []
    c = 6
    for a in range(6, 100):
        for b in range(a, 100):
            # 求三个数的最小公倍数作为分母
            lcm_int = lcm_of_three(a, b, c)
            # 如果分母相同的情况下， 分子之和相等。 那么 a b 则为解
            if lcm_int / a + lcm_int / b == lcm_int / c:
                solutions.append((a, b))

    for solution in solutions:
        print(solution[0], solution[1])Copy

7 42
8 24
9 18
10 15
12 12Copy

solutions = []

for a in range(6, 100):
    for b in  range(a, 100):
        if (a + b) * 6 == a * b:
            solutions.append((a, b))

width = len(str(max(map(max, solutions))))

for a, b in solutions:
    print(f"a = {a:{width}}, b = {b:{width}}")Copy

a =  7, b = 42
a =  8, b = 24
a =  9, b = 18
a = 10, b = 15
a = 12, b = 12Copy

    # 使用 abs 函数和一个小的误差范围来比较两个浮点数
    if abs(1/a + 1/b - 1/6) < 1e-6:
        solutions.append((a, b))Copy