2014 AMC 10B Problems/Problem 17

Problem 17

What is the greatest power of $2$ that is a factor of $10^{1002} - 4^{501}$?

$\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}$

Solution 1

We begin by factoring the $2^{1002}$ out. This leaves us with $5^{1002} - 1$.

We factor the difference of squares, leaving us with $(5^{501} - 1)(5^{501} + 1)$. We note that all even powers of $5$ more than two end in ...$625$. Also, all odd powers of five more than $2$ end in ...$125$. Thus, $(5^{501} + 1)$ would end in ...$126$ and thus would contribute one power of two to the answer, but not more.

We can continue to factor $(5^{501} - 1)$ as a difference of cubes, leaving us with $(5^{167} - 1)$ times an odd number (Notice that the other number is $5^{334} + 5^{167} + 1$. The powers of $5$ end in $5$, so the two powers of $5$ will end with $0$. Adding $1$ will make it end in $1$. Thus, this is an odd number). $(5^{167} - 1)$ ends in ...$124$, contributing two powers of two to the final result.

Or we can see that $(5^{501} - 1)$ ends in $124$, and is divisible by $2$ only. Still that's $2$ powers of $2$.

Adding these extra $3$ powers of two to the original $1002$ factored out, we obtain the final answer of $\textbf{(D) } 2^{1005}$.

Solution 2

First, we can write the expression in a more primitive form which will allow us to start factoring. \[10^{1002} - 4^{501} = 2^{1002} \cdot 5^{1002} - 2^{1002}\] Now, we can factor out $2^{1002}$. This leaves us with $5^{1002} - 1$. Call this number $N$. Thus, our final answer will be $2^{1002+k}$, where $k$ is the largest power of $2$ that divides $N$. Now we can consider $N \pmod{16}$, since $k \le 4$ by the answer choices.

Note that \begin{align*} 5^1 &\equiv 5 \pmod{16} \\ 5^2 &\equiv 9 \pmod{16} \\ 5^3 &\equiv 13 \pmod{16} \\ 5^4 &\equiv 1 \pmod{16} \\ 5^5 &\equiv 5 \pmod{16} \\ &\: \: \qquad \vdots \end{align*} The powers of $5$ cycle in $\mod{16}$ with a period of $4$. Thus, \[5^{1002} \equiv 5^2 \equiv 9 \pmod{16} \implies 5^{1002} - 1 \equiv 8 \pmod{16}\] This means that $N$ is divisible by $8= 2^3$ but not $16 = 2^4$, so $k = 3$ and our answer is $2^{1002 + 3} =\boxed{\textbf{(D)}\: 2^{1005}}$.


See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS