461 + 462 + 463 + 464 + 465 is divisible by :

When every term of a sum is a power of the same base with consecutive exponents, factor out the smallest power first — basem + base(m+1) + ... becomes basem × (1 + base + base2 + ...). If a number shares no common factor with the base, it divides the whole sum exactly when it divides just that bracketed value.

1. Factor out the smallest power, 461, from the sum: 461 + 462 + 463 + 464 + 465 = 461 × (1 + 4 + 42 + 43 + 44).

2. Evaluate the bracket: 1 + 4 + 16 + 64 + 256 = 341.

3. So the whole expression equals 461 × 341.

4. 461 is a power of 2, so it shares no common factor with 3, 11, 17, or 19; divisibility of the product by any of these therefore depends only on whether that number divides 341.

5. Checking 341 against each option shows it is exactly divisible by 11 (341 = 11 × 31) but not by 3, 17, or 19 — so the whole expression is divisible by 11.

Cross-check using modular arithmetic, independent of the factoring above: the powers of 4 modulo 11 cycle with period 5, giving remainders 4, 5, 9, 3, 1. Across the exponents 61 to 65 the corresponding remainders are exactly this same cycle (4, 5, 9, 3, 1), and 4 + 5 + 9 + 3 + 1 = 22, itself a multiple of 11 — confirming the same result independently.

So 461 + 462 + 463 + 464 + 465 is exactly divisible by 11 — not by 3, 17, or 19.