2015 Countdown Problem #8: Factoring 2015 Squared

If No. = a^p × b^q × c^r then distinct divisors/factors of No are (p + 1)(q + 1)(r + 1)

2015^² = 5^2×13^2×31^2

distinct divisors/factors of 2015² = (2 + 1)(2 + 1)(2 + 1) = 27

If $N=a_1^{\alpha_{1}}+a_2^{\alpha_{2}}+a_3^{\alpha_{3}}...+a_n^{\alpha_{n}}$ where $\alpha_{i}$ is a prime factor of $N$ , the number of factor' $N$ is $(\alpha_{1})(\alpha_{2})(\alpha_{3})...(\alpha_{n})$

As $2015=13^2 \times 31^2 \times 5^2 \rightarrow$ it has $3 \times 3 \times 3=27$ divisors

$(2015)^2=5^2×(13)^2×(31)^2$

Case 0: Number of factors being the product of 0 prime factors: 1 (1)

Case 1: Number of factors being the product of 1 prime factor: 3 (5, 13, 31)

Case 2: Number of factors being the product of 2 prime factors: 3+3C2=6 $(5^2,(13)^2,(31)^2,5×13,5×31,13×31)$

Case 3: Number of factors being the product of 3 prime factors: 1+3×2=7 $(5×13×31,5^2×13,5^2×31,(13)^2×5,(13)^2×31,(31)^2×5,(31)^2×13)$

Case 4: Number of factors being the product of 4 prime factors: 3+3=6 $(5^2×13×31,5×(13)^2×31,5×13×(31)^2,5^2×(13)^2 ,(13)^2×(31)^2,5^2×(31)^2)$

Case 5: Number of factors being the product of 5 prime factors: 3 $(5^2×(13)^2×31,5^2×13×(31)^2,5×(13)^2×(31)^2)$

Case 6: Number of factors being the product of 6 prime factors: 1