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2014 AMC 10B Problems/Problem 17: Difference between revisions

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<math>\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}</math>
<math>\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}</math>


==Solution==
==Solution 1==


We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>.  
We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>.  
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Adding these extra <math>3</math> powers of two to the original <math>1002</math> factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>.
Adding these extra <math>3</math> powers of two to the original <math>1002</math> factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>.
==Solution 2==
First, we can write the expression in a more primitive form which will allow us to start factoring.
<cmath>10^{1002} - 4^{501} = 2^{1002} \cdot 5^{1002} - 2^{1002}</cmath>
Now, we can factor out <math>2^{1002}</math>. This leaves us with <math>5^{1002} - 1</math>. Call this number <math>N</math> Thus, our final answer will be <math>2^{1002+k}</math>, where <math>k</math> is the largest power of <math>2</math> that divides <math>N</math>. Now we can consider <math>N \pmod{16}</math>, since <math>k \le 4</math> by the answer choices.
Note that
<cmath>\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*}</cmath>


==See Also==
==See Also==
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 02:06, 23 February 2014

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. $(5^{167} - 1)$ ends in ...$124$, contributing two powers of two to the final result.

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*}

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
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

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