If N = 3^{12} - 1, what is the sum of all the distinct prime factors of N? — GMAT

If $N = 3^{12} - 1$ , what is the sum of all the distinct prime factors of $N$ ?

Answer: 100

The sum of all the distinct prime factors of

N

is 100.

Factoring

3^{12} - 1

via difference of squares and sum of cubes yields

(3^6 - 1)(3^6 + 1) = (2^3 \times 7 \times 13)(2 \times 5 \times 73) = 2^4 \times 5 \times 7 \times 13 \times 73

. The distinct prime factors are 2, 5, 7, 13, and 73, which sum to 100.

Decompose

N = 3^{12} - 1

using the difference of squares identity

a^2 - b^2 = (a - b)(a + b)

N = (3^6 - 1)(3^6 + 1)

Breaking down large powers of integers into products of smaller terms allows for systematic prime factor determination.

Completely factor the term

(3^6 - 1)

3^6 - 1 = (3^3 - 1)(3^3 + 1) = 26 \times 28 = (2 \times 13) \times (2^2 \times 7) = 2^3 \times 7 \times 13

Applying the difference of squares identity iteratively converts the term into small arithmetic integers with obvious prime factorizations.

Completely factor the term

(3^6 + 1)

using the sum of cubes identity

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

3^6 + 1 = (3^2)^3 + 1 = (3^2 + 1)((3^2)^2 - 3^2 + 1) = 10 \times 73 = 2 \times 5 \times 73

The factor 73 is prime because it is not divisible by any prime numbers less than or equal to

\sqrt{73} \approx 8.54

(namely 2, 3, 5, and 7).

Combine the component prime factorizations to list all distinct prime factors of

N

N = 2^4 \times 5 \times 7 \times 13 \times 73

, so the set of distinct prime factors is \{2, 5, 7, 13, 73\}.

Each prime base is included exactly once regardless of its exponent.

Sum the distinct prime factors.

2 + 5 + 7 + 13 + 73 = 100

Adding the unique prime factors yields the requested value.

Prime Factorization using Algebraic Polynomial Identities

Estimated Time:2m 30s

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