Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 6)
 
(42 intermediate revisions by 14 users not shown)
Line 1: Line 1:
−
==Problem 17==
+
==Problem==
 
What is the greatest power of <math>2</math> that is a factor of <math>10^{1002} - 4^{501}</math>?
 
What is the greatest power of <math>2</math> that is a factor of <math>10^{1002} - 4^{501}</math>?
  
Line 8: Line 8:
 
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>.  
  
−
We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that all even powers of 5 more than two end in ...<math>625</math>. Also, all odd powers of five more than 2 end in  ...<math>125</math>. Thus, <math>(5^{501} + 1)</math> would end in ...<math>126</math> and thus would contribute one power of two to the answer, but not more.  
+
We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that all even powers of <math>5</math> more than two end in ...<math>625</math>. Also, all odd powers of five more than <math>2</math> end in  ...<math>125</math>. Thus, <math>(5^{501} + 1)</math> would end in ...<math>126</math> and thus would contribute one power of two to the answer, but not more.  
  
−
We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number. <math>(5^{167} - 1)</math> ends in ...<math>124</math>, contributing two powers of two to the final result.
+
We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number (Notice that the other number is <math>5^{334} + 5^{167} + 1</math>. The powers of <math>5</math> end in <math>5</math>, so the two powers of <math>5</math> will end with <math>0</math>. Adding <math>1</math> will make it end in <math>1</math>. Thus, this is an odd number). <math>(5^{167} - 1)</math> ends in ...<math>124</math>, contributing two powers of two to the final result.
 +
 
 +
Or we can see that <math>(5^{501} - 1)</math> ends in <math>124</math>, and is divisible by <math>2</math> only. Still that's <math>2</math> powers of <math>2</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>.
 
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>.
Line 18: Line 20:
 
First, we can write the expression in a more primitive form which will allow us to start factoring.
 
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>
 
<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.  
+
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
 
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>
 
<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>
 +
The powers of <math>5</math> cycle in <math>\mod{16}</math> with a period of <math>4</math>. Thus, <cmath>5^{1002} \equiv 5^2 \equiv 9 \pmod{16} \implies 5^{1002} - 1 \equiv 8 \pmod{16}</cmath>
 +
This means that <math>N</math> is divisible by <math>8= 2^3</math> but not <math>16 = 2^4</math>, so <math>k = 3</math> and our answer is <math>2^{1002 + 3} =\boxed{\textbf{(D)}\: 2^{1005}}</math>.
 +
 +
==Solution 3==
 +
Convert <math>4^{501}=2^{1002}</math>. We can factor out <math>2^{1002}</math> to get that <math>\nu_2(10^{1002}-2^{1002})=1002+\nu_2(5^{1002}-1)</math>. Using the adjusted Lifting The Exponent lemma (<math>\nu_2(a^n-b^n)=\nu_2(n)+\nu_2(a^2-b^2)-1</math> for all even <math>n</math> and odd <math>a,b</math>), we get that the answer is <math>2^{1002+\nu_2(1002)+\nu_2(24)-1}=2^{1002+1+3-1}=\boxed{\textbf{(D)}2^{1005}}</math>
 +
 +
==Solution 4==
 +
 +
Factor out <math>2^{1002}</math> to get <math>2^{1002}(5^{1002} - 1)</math>. Since <math>5^{1002}-1\equiv 3^{1002}-1\equiv (9)^{501}-1\equiv 1^{501}-1\equiv 0\pmod{8}</math>, but <math>5^{1002}-1\equiv 11^{1002}-1\equiv 121\cdot (11^4)^{250} - 1\equiv 121 - 1\equiv 8 \pmod{16}</math>, <math>5^{1002}-1</math> has 3 factors of 2. Hence <math>2^{1002 + 3} =\boxed{2^{1005}}</math> is the largest power of two which divides the given number
 +
 +
==Solution 5==
 +
 +
Like Solution 1, factor out <math>2^{1002}</math> to get <math>(2^{1002})(5^{501}-1)(5^{501}+1)</math>. Using engineer's induction, we observe that for any positive integer <math>5^n</math> (where <math>n</math> is an odd positive integer), it appears that the least even numbers directly above and below <math>n</math> in value must contain a maximum multiple of <math>4</math> and a maximum multiple of <math>2</math>. Hence, the answer is <math>2^{1002+2+1}</math> which is <math>\boxed{\textbf{(D)} 2^{1005}}</math> .
 +
 +
Proof;
 +
 +
For all integers <math>x</math> where <math>x=5^{n}</math> where n is an odd integer, <math>x</math> must end in <math>125.</math>
 +
Thus, we find that <math>x-1</math> and <math>x+1</math> respectively end in <math>124</math> and <math>126.</math>
 +
 +
Case <math>1</math> : <math>x-1</math>
 +
 +
We know that this number takes the form <math>abcde... 124</math> where <math>abcde...</math> is an integer that ends in <math>124</math>. Because <math>abcde...</math> is a multiple of <math>4</math> times an even number <math>e</math> while <math>124</math> is <math>4(31)</math>, we find that <math>X-1</math> must be <math>4e + 4 \cdot 31 = 4(e+31) = 4o</math> where <math>o</math> is an odd number
 +
 +
Case <math>2</math> : <math>x+1</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 <math>2</math> while <math>x+1</math> must have a maximum of <math>1</math> factor of <math>2</math>.
 +
 +
 +
 +
 +
~ShangJ2
 +
 +
==Solution 6==
 +
 +
Using difference of squares, we get <math>\left(10^{501}-2^{501}\right) \left(10^{501}+2^{501}\right)</math>. Factoring a <math>2^{501}</math> out, we get <math>\left(2^{501}\right) \left(5^{501}-1\right) \left(2^{501}\right) \left(5^{501}+1\right)</math>, and grouping like terms give <math>\left(2^{1002}\right) \left(5^{501}-1\right) \left(5^{501}+1\right)</math>.
 +
 +
 +
Then, you would go ahead and innocently choose <math>\textbf{(A) } 2^{1002}</math>, right? No! Note that <math>5^n</math>, where <math>n</math> is any odd integer greater than or equal to <math>3</math>, it always ends in <math>125</math>. So, <math>5^{501}+1</math> ends in <math>126</math> and <math>5^{501}-1</math> ends in <math>124</math>, so they add up to an extra three <math>2</math>'s. Therefore, the answer is actually <math>2^{1002+3}=\boxed{\textbf{(D) } 2^{1005}}</math>.
 +
 +
~MrThinker
 +
 +
==Video Solution==
 +
https://youtu.be/aCtvD8nitgg
 +
 +
~savannahsolver
 +
  
 
==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}}

Latest revision as of 18:47, 11 August 2023

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 an 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 $4e + 4 \cdot 31 = 4(e+31) = 4o$ 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

Solution 6

Using difference of squares, we get $\left(10^{501}-2^{501}\right) \left(10^{501}+2^{501}\right)$. Factoring a $2^{501}$ out, we get $\left(2^{501}\right) \left(5^{501}-1\right) \left(2^{501}\right) \left(5^{501}+1\right)$, and grouping like terms give $\left(2^{1002}\right) \left(5^{501}-1\right) \left(5^{501}+1\right)$.


Then, you would go ahead and innocently choose $\textbf{(A) } 2^{1002}$, right? No! Note that $5^n$, where $n$ is any odd integer greater than or equal to $3$, it always ends in $125$. So, $5^{501}+1$ ends in $126$ and $5^{501}-1$ ends in $124$, so they add up to an extra three $2$'s. Therefore, the answer is actually $2^{1002+3}=\boxed{\textbf{(D) } 2^{1005}}$.

~MrThinker

Video Solution

https://youtu.be/aCtvD8nitgg

~savannahsolver


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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_17&oldid=196811"