2014 AMC 10B Problem 17

Write both terms with base $2$: $10^{1002} = 2^{1002} \cdot 5^{1002}$ and $4^{501} = 2^{1002}$. Hence
$$10^{1002} - 4^{501} = 2^{1002}\left(5^{1002} - 1\right).$$

Now count the factors of $2$ in $5^{1002} - 1$. By difference of squares,
$$5^{1002} - 1 = \left(5^{501} - 1\right)\left(5^{501} + 1\right).$$

• $5^{501} + 1$: since $5 \equiv 1 \pmod 4$, $5^{501} \equiv 1 \pmod 4$, so $5^{501} + 1 \equiv 2 \pmod 4$. It is divisible by $2$ but not by $4$: exactly one factor of $2$.
• $5^{501} - 1 = (5 - 1)\left(5^{500} + 5^{499} + \cdots + 5 + 1\right)$. The second factor is a sum of $501$ odd numbers, hence odd. So $5^{501} - 1$ has exactly the two factors of $2$ from $5 - 1 = 4$.

Therefore $5^{1002} - 1$ has exactly $2^{1 + 2} = 2^3$, and the whole expression has $2^{1002 + 3} = 2^{1005}$.

The answer is $\boxed{\textbf{(D)}\ 2^{1005}}$.

"Greatest power of $2$" asks for the exact exponent of $2$ in a factorization, so factor as far as possible and count in each piece. Factoring out the obvious common power is only the first step; the leftover $5^{1002} - 1$ is even and must be examined too. The $a^n - 1$ factorization plus a mod-$4$ check is the standard way to count 2s in such expressions.

Work mod $16$: $5^2 = 25 \equiv 9$ and $9^2 = 81 \equiv 1 \pmod{16}$, so $5^{1002} = 25^{501} \equiv 9^{501} = 9^{500} \cdot 9 \equiv 9 \pmod{16}$. Thus $5^{1002} - 1 \equiv 8 \pmod{16}$: divisible by $8$ but not $16$, confirming exactly three extra factors of $2$.

• Stopping at 2^1002 after factoring it out, forgetting that 5^1002 - 1 is itself even and contributes three more factors of 2; this gives choice (A).
• Assuming $5^{1002} - 1$ contributes only one factor of $2$ because "even minus one is odd, odd minus one is even," giving $2^{1003}$.