Difference between revisions of "2014 AMC 10B Problems/Problem 17"

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Problem

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 $2^{1002 + 3} =\boxed{\textbf{(D)}\: 2^{1005}}$ .

Solution 3

Convert $4^{501}=2^{1002}$ . We can factor out $2^{1002}$ to get that $\nu_2(10^{1002}-2^{1002})=1002+\nu_2(5^{1002}-1)$ . Using the adjusted Lifting The Exponent lemma ( $\nu_2(a^n-b^n)=\nu_2(n)+\nu_2(a^2-b^2)-1$ for all even $n$ and odd $a,b$ ), we get that the answer is $2^{1002+\nu_2(1002)+\nu_2(24)-1}=2^{1002+1+3-1}=\boxed{\textbf{(D)}2^{1005}}$

Solution 4

Factor out $2^{1002}$ to get $2^{1002}(5^{1002} - 1)$ . Since $5^{1002}-1\equiv 3^{1002}-1\equiv (9)^{501}-1\equiv 1^{501}-1\equiv 0\pmod{8}$ , but $5^{1002}-1\equiv 11^{1002}-1\equiv 121\cdot (11^4)^{250} - 1\equiv 121 - 1\equiv 8 \pmod{16}$ , $5^{1002}-1$ has 3 factors of 2. Hence $2^{1002 + 3} =\boxed{2^{1005}}$ is the largest power of two which divides the given number

Solution 5

Like Solution 1, factor out $2^{1002}$ to get $(2^{1002})(5^{501}-1)(5^{501}+1)$ . Using engineer's induction, we observe that for any positive integer $5^n$ (where $n$ is an odd positive integer), it appears that the least even numbers directly above and below $n$ in value must contain a maximum multiple of $4$ and a maximum multiple of $2$ . Hence, the answer is $2^{1002+2+1}$ which is $\boxed{\textbf{(D)} 2^{1005}}$ .

Proof ;

For all integers $x$ where $x=5^{n}$ where n is an odd integer, $x$ must end in $125.$ Thus, we find that x-1 and x+1 respectively end in 124 and 126. Case $1$ : $x-1$

We know that this number takes the form abcde... 124 where abcde is a integer that ends in 124. Because abcde is a multiple of $4$ times an even number $e$ while $124$ is $4(31)$ , we find that $X-1$ must be $4e + 4*31 = 4(e+31) = 4o$ where $o$ is an odd number

Case $2$ : $x+1$

We know that this number $fghijk...126$ ends. Because it is $2$ more than the number $x-1$ , which is a multiple of $4$ , we find $x+1 = 4o + 2$ which is an even number that is not divisible by $4$ . Thus, it must have a maximum of $1$ multiple of $2$ .

This means that for any number $x$ being in the form $5^{n}$ where $n$ is an odd integer, $x-1$ must have a maximum of $2$ factors of 2 while $x+1$ must have a maximum of $1$ factor of $2$ .

~ShangJ2

Video Solution

https://youtu.be/aCtvD8nitgg

~savannahsolver

@@ Line 48: / Line 48: @@
 Case <math>2</math> : <math>x+1</math>
-we know that this number fghijk... <math>126</math> ends. Because it is <math>2</math> more than the number <math>x-1</math>, which is a multiple of <math>4</math>, we find <math>x+1 = 4o + 2</math> which is an even number that is not divisible by <math>4</math>. Thus, it must have a maximum of <math>1</math> multiple of <math>2</math>.
+We know that this number <math>fghijk...126</math> ends. Because it is <math>2</math> more than the number <math>x-1</math>, which is a multiple of <math>4</math>, we find <math>x+1 = 4o + 2</math> which is an even number that is not divisible by <math>4</math>. Thus, it must have a maximum of <math>1</math> multiple of <math>2</math>.
 This means that for any number <math>x</math> being in the form <math>5^{n}</math> where <math>n</math> is an odd integer, <math>x-1</math> must have a maximum of <math>2</math> factors of 2 while <math>x+1</math> must have a maximum of <math>1</math> factor of <math>2</math>.

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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