Powers and Roots Practice Test 4

Concept:Factor out the common power of $2$, then count the extra factor of $2$ left inside the bracket.Explanation:$10^{2008} = (2 \times 5)^{2008} = 2^{2008} \times 5^{2008}$.So, $2^{2008} + 10^{2008} = 2^{2008} + 2^{2008} \times 5^{2008}$.Taking $2^{2008}$ common gives $2^{2008}(1 + 5^{2008})$.Since $5^{2008}$ is odd, $1 + 5^{2008}$ is even, so it contributes at least one more factor of $2$.Now check how many factors of $2$ are in $1 + 5^{2008}$.For any odd number squared, $5^2 \equiv 1 \pmod{8}$.Thus, $5^{2008} = (5^2)^{1004} \equiv 1 \pmod{8}$.Therefore, $1 + 5^{2008} \equiv 1 + 1 = 2 \pmod{8}$.This means $1 + 5^{2008}$ has exactly one factor of $2$, not more.So the total power of $2$ dividing the expression is $2008 + 1 = 2009$.Answer:The largest power of $2$ that divides $(2^{2008} + 10^{2008})$ is $2^{2009}$ (Option B).