A natural number n has nine divisors, while n-1 has eight divisors. What are the smallest and the second smallest of the natural numbers satisfying the conditions?

A natural number n has nine divisors, while n-1 has eight divisors. What are the smallest and the second smallest of the natural numbers satisfying the conditions?

According to the formula of the number of divisors, we can know: ① when n = an, that is, n has only one prime factor, N + 1 = 9, so n = 8, so the smallest n = 28 = 256, n-1 = 255 = 3 × 5 × 17, there are exactly (1 + 1) × (1 + 1) × (1 + 1) = 8 divisors, which is in line with the meaning of the problem; ② when n = an × BM, that is, n has two prime factors, (n + 1) (M + 1) = 9, so n = m = 2, so the smallest n = 22 × 32 = 36, n-1 = 35 = 5 × 7, there are (1 + 1) × The second smallest n = 22 × 52 = 50, n-1 = 49 = 7 × 7, has (1 + 1) × (1 + 1) = 4 divisors, which is not in line with the meaning of the topic; the third smallest n = 22 × 72 = 196, n-1 = 195 = 3 × 5 × 13, has (1 + 1) × (1 + 1) × (1 + 1) = 8 divisors, which is in line with the meaning of the topic; to sum up, the smallest n is 196, and the second smallest is 256