Find the sum of even divisors of 4096 .

Concept: When a number is a pure power of 2, N = 2a, its only divisors are 20, 21, …, 2a. Among these, only 20 = 1 is odd — every other divisor is even. So the even divisors form the geometric progression 21, 22, …, 2a, with first term 2 and common ratio 2, and the sum of n terms of a GP is a(rn − 1)/(r − 1).

Application:

1. Factorize 4096: 4096 = 212 (since 212 = 4096).

2. 4096 has 13 divisors in total (20 through 212); exactly one of them, 20 = 1, is odd, so the remaining 12 divisors — 21, 22, …, 212 — are all even.

3. These 12 even divisors form a geometric progression with first term a = 2, common ratio r = 2, and n = 12 terms.

4. Sum = a(rn − 1)/(r − 1) = 2 × (212 − 1)/(2 − 1) = 2 × 4095 = 8190.

Cross-check: The sum of ALL divisors of 212 (odd + even) is (213 − 1)/(2 − 1) = 8191. Removing the one odd divisor (1) gives 8191 − 1 = 8190 — the same result by an independent route.

So the sum of the even divisors of 4096 is 8190.