1. **State the problem:** We want to find a number by which the sum $4^{41} + 4^{42} + 4^{43}$ is divisible. 2. **Rewrite the expression:** Notice that $4^{42} = 4^{41} \times 4$ and $4^{43} = 4^{41} \times 4^2$. 3. **Factor out the common term:** $$4^{41} + 4^{42} + 4^{43} = 4^{41} + 4^{41} \times 4 + 4^{41} \times 4^2 = 4^{41}(1 + 4 + 16)$$ 4. **Simplify inside the parentheses:** $$1 + 4 + 16 = 21$$ 5. **Final factorization:** $$4^{41} \times 21$$ 6. **Interpretation:** Since the sum factors as $4^{41} \times 21$, it is divisible by both $4^{41}$ and $21$. 7. **Answer:** The sum $4^{41} + 4^{42} + 4^{43}$ is divisible by $4^{41}$ and also by $21$. **Therefore, the sum is divisible by $4^{41}$ and $21$.