we can write $$2^{16} - 1$$ as $$16^4 - 1^4$$ and we know $$a^n - b^n$$ is always divisible by (a-b) and (a+b) if n is even hence $$16^4 - 1^4$$ will always be divisible by (16-1) and (16+1) and hence the answer is 17
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use