Difference between revisions of "2014 AMC 10B Problems/Problem 17"
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==Solution 1== | ==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>. |
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 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. | ||
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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> | 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>. | 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>. | ||
+ | PLS HELP! THIS MAKES NO SENSE! IT SOUNDS LIKE JIBBERISH!!! | ||
==Solution 3== | ==Solution 3== | ||
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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 | 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 4== | + | ==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 | ~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== | ==Video Solution== |
Revision as of 10:51, 1 June 2024
Contents
Problem
What is the greatest power of that is a factor of
?
Solution 1
We begin by factoring the out. This leaves us with
.
We factor the difference of squares, leaving us with . We note that all even powers of
more than two end in ...
. Also, all odd powers of five more than
end in ...
. Thus,
would end in ...
and thus would contribute one power of two to the answer, but not more.
We can continue to factor as a difference of cubes, leaving us with
times an odd number (Notice that the other number is
. The powers of
end in
, so the two powers of
will end with
. Adding
will make it end in
. Thus, this is an odd number).
ends in ...
, contributing two powers of two to the final result.
Or we can see that ends in
, and is divisible by
only. Still that's
powers of
.
Adding these extra powers of two to the original
factored out, we obtain the final answer of
.
Solution 2
First, we can write the expression in a more primitive form which will allow us to start factoring.
Now, we can factor out
. This leaves us with
. Call this number
. Thus, our final answer will be
, where
is the largest power of
that divides
. Now we can consider
, since
by the answer choices.
Note that
The powers of
cycle in
with a period of
. Thus,
This means that
is divisible by
but not
, so
and our answer is
.
PLS HELP! THIS MAKES NO SENSE! IT SOUNDS LIKE JIBBERISH!!!
Solution 3
Convert . We can factor out
to get that
. Using the adjusted Lifting The Exponent lemma (
for all even
and odd
), we get that the answer is
Solution 4
Factor out to get
. Since
, but
,
has 3 factors of 2. Hence
is the largest power of two which divides the given number
Solution 5
Like Solution 1, factor out to get
. Using engineer's induction, we observe that for any positive integer
(where
is an odd positive integer), it appears that the least even numbers directly above and below
in value must contain a maximum multiple of
and a maximum multiple of
. Hence, the answer is
which is
.
Proof;
For all integers where
where n is an odd integer,
must end in
Thus, we find that
and
respectively end in
and
Case :
We know that this number takes the form where
is an integer that ends in
. Because
is a multiple of
times an even number
while
is
, we find that
must be
where
is an odd number
Case :
We know that this number ends. Because it is
more than the number
, which is a multiple of
, we find
which is an even number that is not divisible by
. Thus, it must have a maximum of
multiple of
.
This means that for any number being in the form
where
is an odd integer,
must have a maximum of
factors of
while
must have a maximum of
factor of
.
~ShangJ2
Solution 6
Using difference of squares, we get . Factoring a
out, we get
, and grouping like terms give
.
Then, you would go ahead and innocently choose , right? No! Note that
, where
is any odd integer greater than or equal to
, it always ends in
. So,
ends in
and
ends in
, so they add up to an extra three
's. Therefore, the answer is actually
.
~MrThinker
Video Solution
~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.