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4095 equals $2^{12} - 1$ . What is the sum of its distinct prime factors?

Competition practiceNumber System, HCF & LCM

A28
B31
C25
D40

Step-by-Step Solution

2^{12} - 1 = (2^{6} - 1) (2^{6} + 1) = 63 \times 65 = 3^{2} \times 7 \times 5 \times 13

. Distinct primes 3, 5, 7, 13 add to 28. (31 counts 3 twice.)

Answer: A) 28

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