It is being given that (232 + 1) is completely divisible by a whole number.…

Concept: The sum-of-cubes identity states that for any values a and b, a3 + b3 factors as (a + b)(a2 - ab + b2). Whenever a number divides the factor (a + b), it automatically divides the entire product a3 + b3, because that product is simply (a + b) multiplied by another whole number.

Application:

1. Let N be the whole number that divides (232 + 1). Set x = 232, so the given fact becomes: N divides (x + 1).

2. Rewrite (296 + 1) in terms of x: since 96 = 32 x 3, 296 = (232)3 = x3. So (296 + 1) = x3 + 1.

3. Apply the sum-of-cubes identity to x3 + 1: x3 + 1 = (x + 1)(x2 - x + 1).

4. Since N divides (x + 1), and (x + 1) is a factor of the product (x + 1)(x2 - x + 1) = x3 + 1, N must divide the whole product. So N divides (296 + 1).

Cross-check: The other three expressions have no such forced factor relationship with (232 + 1). For a concrete check, 232 + 1 = 4294967297 = 641 x 6700417, so N = 641 is one valid whole-number divisor. Testing N = 641 against the other options: 216 + 1 = 65537 and 216 - 1 = 65535 are not multiples of 641, and 7 x 223 = 58720256 is not a multiple of 641 either (641 is odd, so it shares no factor with a pure power of 2). Only (296 + 1) comes out as a guaranteed multiple.

Therefore, the number completely divisible by the same whole number is (296 + 1).