Difference between revisions of "2011 AIME I Problems/Problem 11"
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− | == Problem == | + | ==Problem== |
− | |||
− | == Solution == | + | Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by 1000. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000. |
− | Note that the | + | |
− | <cmath>S=2^0+2^1+2^2+2^3+2^4...+2^{102}\equiv 2^{103}-1\equiv 8-1\equiv \boxed{007}\mod 1000</cmath> | + | == Solution 1== |
+ | Note that <math>x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}</math> and <math>x \equiv y \pmod{8}</math>. So we must find the first two integers <math>i</math> and <math>j</math> such that <math>2^i \equiv 2^j \pmod{125}</math> and <math>2^i \equiv 2^j \pmod{8}</math> and <math>i \neq j</math>. Note that <math>i</math> and <math>j</math> will be greater than 2 since remainders of <math>1, 2, 4</math> will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that <math>2^{100}\equiv 1\pmod{125}</math> (see [[Euler's Totient Theorem|Euler's theorem]]) and <math>2^0,2^1,2^2,\ldots,2^{99}</math> are all distinct modulo 125 (proof below). Thus, <math>i = 103</math> and <math>j =3</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation: | ||
+ | <cmath> | ||
+ | S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}. | ||
+ | </cmath> | ||
+ | To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of <math>2^{20}\equiv 1\pmod{125}</math> or <math>2^{50}\equiv 1\pmod{125}</math>. However, writing <math>2^{10}\equiv 25 - 1\pmod{125}</math>, we can easily verify that <math>2^{20}\equiv -49\pmod{125}</math> and <math>2^{50}\equiv -1\pmod{125}</math>, giving us the needed contradiction. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Notice that our sum of remainders looks like <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots.</cmath> We want to find the remainder of <math>S</math> upon division by <math>1000.</math> Since <math>1000</math> decomposes into primes as <math>1000 = 2^3 \cdot 5^3</math>, we can check the remainders of <math>S</math> modulo <math>2^3</math> and modulo <math>5^3</math> separately. | ||
+ | |||
+ | Checking <math>S</math> modulo <math>2^3</math> is easy, so lets start by computing the remainder of <math>S</math> upon division by <math>5^3.</math> To do this, let's figure out when our sequence finally repeats. | ||
+ | Notice that since the remainder when dividing any term of <math>S</math> (after the third term) by <math>1000</math> will be a multiple of <math>2^3</math>, when this summation finally repeats, the first term to repeat will be ''not'' be <math>1</math> since <math>2^3 \nmid 1.</math> Instead, the first term to repeat will be <math>2^3</math>, and then the sequence will continue once again <math>2^4, 2^5, \cdots.</math> | ||
+ | |||
+ | Now, to compute <math>S</math> modulo <math>125</math>, we want to find the least positive integer <math>d</math> such that <math>2^d \equiv 1 \pmod {125}</math> since then <math>d</math> will just be the number of terms of <math>S</math> (after the third term!) before the sequence repeats. In other words, our sequence will be of the form <math>S = 1 + 2 + 4 + \left(2^3 + 2^4 + \cdots + 2^{2 + d}\right)</math> and then we will have <math>2^{d + 3} \equiv 2^3 \pmod {125}</math>, and the sequence will repeat from there. Here, <math>d</math> simply represents the order of <math>2</math> modulo <math>125</math>, denoted by <math>d = \text{ord}_{125}(2).</math> To begin with, we'll use a well-known property of the order to get a bound on <math>d.</math> | ||
+ | |||
+ | Since <math>\gcd(2, 125) = 1</math> and <math>\phi(125) = 100</math>, we know by Euler's Theorem that <math>2^{100} \equiv 1 \pmod {125}.</math> However, we do not know that <math>100</math> is the ''least'' <math>d</math> satisfying <math>2^d \equiv 1 \pmod {125}.</math> Nonetheless, it is a well known property of the order that <math>\text{ord}_{125}(2) = d | \phi(125) = 100.</math> Therefore, we can conclude that <math>d</math> must be a positive divisor of <math>100.</math> | ||
+ | |||
+ | Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say <math>\mod 5.</math> Clearly, we must have that <math>2^d \equiv 1 \pmod 5.</math> Since <math>2^4 \equiv 1 \pmod 5</math> and powers of two will then cycle every four terms, we know that <math>2^d \equiv 1 \pmod 5 \iff 4 | d.</math> Combining this relation with <math>d | 100</math>, it follows that <math>d \in \{4, 20, 100\}.</math> | ||
+ | |||
+ | Now, it is trivial to verify that <math>d \ne 4.</math> In addition, we know that <cmath>2^{20} = \left(2^{10}\right)^2 = \left(1024\right)^2 \equiv 24^2 = 576 \not\equiv 1 \pmod {125}.</cmath> Therefore, we conclude that <math>d \ne 20.</math> Hence, we must have <math>d = 100.</math> (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly). | ||
+ | |||
+ | Now, since we have found <math>d = 100</math>, we know that <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots + 2^{102}.</cmath> There are two good ways to finish from here: | ||
+ | |||
+ | The first way is to use a trick involving powers of <math>2.</math> Notice that <cmath>S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.</cmath> Certainly <cmath>S = 2^{103} - 1 \equiv -1 \equiv 7 \pmod {8}.</cmath> In addition, since we already computed <math>\text{ord}_{125}(2) = d = 100</math>, we know that <cmath>S = 2^{103} - 1 = 2^{100} \cdot 2^3 - 1 \equiv 2^3 - 1 \equiv 7 \pmod {125}.</cmath> Therefore, since <math>S \equiv 7 \pmod{8}</math> and <math>S \equiv 7 \pmod{125}</math>, we conclude that <math>S \equiv \boxed{007} \pmod {1000}.</math> | ||
+ | |||
+ | The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of <math>S</math> into groups: <cmath>R = 1 + 2 + 4; \quad T = 2^3 + 2^4 + \cdots + 2^{102}.</cmath> It is easy to see that <math>R</math> is congruent to <math>7</math> modulo <math>1000.</math> | ||
+ | |||
+ | Now, for <math>T</math>, notice that there are <math>100</math> terms in the summation, each with a different remainder upon division by <math>125.</math> Since each of these remainders is certainly relatively prime to <math>125</math>, these <math>100</math> remainders correspond to the <math>100</math> positive integers less than <math>125</math> that are relatively prime to <math>125.</math> Therefore, | ||
+ | <cmath>\begin{align*}T &\equiv 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + \cdots + 124 \pmod{125} \\ | ||
+ | &= \left(1 + 2 + 3 + \cdots + 125\right) - \left(5 + 10 + 15 + \cdots 125\right) \\ | ||
+ | &= \frac{125 \cdot 126}{2} - 5\left(1 + 2 + 3 + \cdots 25\right) \\ | ||
+ | &= 125 \cdot 63 - 5\left(\frac{25 \cdot 26}{2}\right) \\ | ||
+ | &= 125\left(63 - 13\right) \\ | ||
+ | &\equiv 0 \pmod{125}.\end{align*}</cmath> | ||
+ | Then, since <math>T</math> is divisible by <math>125</math> and <math>8</math>, it follows that <math>T</math> is divisible by <math>1000.</math> Therefore, <cmath>S = R + T \equiv R \equiv \boxed{007} \pmod{1000}.</cmath> | ||
+ | == Solution 3 == | ||
+ | Obviously, <math>2^i</math> will have to repeat at some point, and our goal is just to find when it repeats. Suppose <math>2^a</math> is the first time the powers of 2 repeat mod 1000, and that it is the same as <math>2^b</math> where <math>b < a.</math> We have | ||
+ | <cmath>2^a \equiv 2^b \mod 1000 \rightarrow 2^a - 2^b \equiv 0 \mod 1000 </cmath> | ||
+ | We can factor out a <math>2^b</math> to get | ||
+ | <cmath>2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000</cmath> | ||
+ | Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since <math>2^{a-b} - 1</math> has to be odd, this necessarily means <math>b \ge 3</math> for <math>8 \div 2^b.</math> This means that 125 has to divide <math>2^{a-b} - 1.</math> We wish to find the minimum <math>a - b</math> such that this is true, or similarly, we just wish to find <math>\text{ord}_{125}(2).</math> Note that by Euler's Totient Theorem, that <math>2^{100} \equiv 1 \mod 125.</math> After checking, and seeing that <math>2^{50} \equiv -1 \mod 125</math>, this directly means that <math>\text{ord}_{125}(2) = 100</math> or <math>a - b = 100.</math> Since <math>b \ge 3</math>, the smallest <math>(a, b)</math> pair is | ||
+ | <math>(103, 3).</math> What this really means is that the sequence of remainders will start repeating at <math>2^{103}</math> or namely that | ||
+ | <math>\{2^0, 2^1, ..., 2^{102}\}</math> are all distinct residues mod 1000. Adding these yields <cmath>2^{103} - 1 \equiv 8 - 1 \equiv \boxed{7} \mod 1000</cmath> | ||
+ | ==Solution 4 (Not rigorous)== | ||
+ | If we write out the remainders when powers of <math>2</math> are divided by <math>1000</math>, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down. | ||
+ | |||
+ | The answer extraction of the problem is equivalent to asking for <math>1+2+4 + \dots + 2^n \pmod{1000}</math>, where <math>2^{n+1}</math> is the first number whose remainder when divided by <math>1000</math> is a repeat. This expression is equivalent to <math>2^{n+1}-1 \pmod{1000}</math>. So our answer will just be the first repeated remainder minus <math>1</math>. | ||
+ | |||
+ | (Everything up to this point has been perfectly rigorous; now, we will destroy this.) | ||
+ | |||
+ | Note that <math>1</math>, <math>2</math> and <math>4</math> cannot be the first repeated remainders, due to the <math>2</math>, <math>4</math> and <math>8</math> divisibility rules, respectively (i.e. <math>2^0</math>, <math>2^1</math> and <math>2^2</math> are the only powers of <math>2</math> that leave a remainder of <math>1</math>, <math>2</math> and <math>4</math>, respectively, when divided by <math>1000</math>.) There is no reason why <math>8</math> could not be the first repeated remainder, so it probably is. Thus, our answer is <math>8-1 = \boxed{7}</math>. (In fact, one can quite easily show that if <math>8</math> reappears at all in the sequence, it must be the first repeated remainder.) | ||
+ | |||
+ | ~ihatemath123 | ||
+ | |||
+ | |||
+ | ==Solution 5 (Very bad idea)== | ||
+ | We can simply list out the last <math>3</math> digits of all the powers of <math>2</math> until we find a pattern. | ||
+ | |||
+ | Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way | ||
+ | |||
+ | <math>001, 002, 004, 008, 016, 032, 064, 128, 256, 512, 024, 048, 096, 192, 384, 768, 536, 072, 144, 288, 576, 152, 304, 608,</math> <math>216, 432, 864, 728, 456, 912, 824, 648, 296, 592, 184, 368, 736, 472, 944, 888, 776, 552, 104, 208, 416, 832, 664, 328,</math> <math>656, 312, 624, 248, 496, 992 \cdots</math> | ||
+ | |||
+ | We can see that after <math>496</math> comes <math>992</math> which can be written as <math>-8 \pmod {1000}</math>. That means that all the terms after <math>8</math> will repeat but negated modulo <math>1000</math>. | ||
+ | Note that <math>n \pmod m + -n \pmod m = 0 \pmod m</math>. As we are looking for the sum of all of the possible values mod <math>1000</math>, the answer is just <math>1 + 2 + 4 = \fbox{007}</math> | ||
+ | |||
+ | ~EvanZ | ||
+ | |||
+ | ==Solution 6 (not rigorous) == | ||
+ | Note that the problem essentially asks for the sum of all <math>x</math> in the residue class of <math>1000</math> that come from a power of <math>2</math>. We can split up this residue class into finding a solution for <math>x</math> in the modular system <cmath>x \equiv 2^n \bmod 8</cmath> <cmath> x \equiv 2^n \bmod 125</cmath> Clearly, if <math>n \geq 3</math>, we get the first equation to be <cmath>x \equiv 0 \bmod 8</cmath> If <math>n \leq 2</math>, we get our residues to be <math>1</math>,<math>2</math>, and <math>4</math> which add to <math>7</math>. Now, we just need to worry about the <math>125</math> mod equation. If we take a look at the powers of <math>2</math> modulo <math>5</math>, we see that we get every residue except for <math>0</math>. If we consider it modulo <math>25</math>, we see that we get every residue except for <math>0,5,10,15,20</math> (ie. all multiples of <math>5</math>). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of <math>5</math> in modulo <math>125</math>. Therefore, we can add all multiples of <math>8</math> from <math>0</math> to <math>1000</math> and subtract the multiples of <math>40</math> to get <math>0 \bmod 1000</math>. Therefore, our final answer is <math>\fbox{7}</math>. | ||
+ | |||
+ | ~Vedoral | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2011|n=I|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category: Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:00, 19 June 2024
Contents
Problem
Let be the set of all possible remainders when a number of the form
,
a nonnegative integer, is divided by 1000. Let
be the sum of the elements in
. Find the remainder when
is divided by 1000.
Solution 1
Note that and
. So we must find the first two integers
and
such that
and
and
. Note that
and
will be greater than 2 since remainders of
will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that
(see Euler's theorem) and
are all distinct modulo 125 (proof below). Thus,
and
are the first two integers such that
. All that is left is to find
in mod
. After some computation:
To show that
are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of
or
. However, writing
, we can easily verify that
and
, giving us the needed contradiction.
Solution 2
Notice that our sum of remainders looks like We want to find the remainder of
upon division by
Since
decomposes into primes as
, we can check the remainders of
modulo
and modulo
separately.
Checking modulo
is easy, so lets start by computing the remainder of
upon division by
To do this, let's figure out when our sequence finally repeats.
Notice that since the remainder when dividing any term of
(after the third term) by
will be a multiple of
, when this summation finally repeats, the first term to repeat will be not be
since
Instead, the first term to repeat will be
, and then the sequence will continue once again
Now, to compute modulo
, we want to find the least positive integer
such that
since then
will just be the number of terms of
(after the third term!) before the sequence repeats. In other words, our sequence will be of the form
and then we will have
, and the sequence will repeat from there. Here,
simply represents the order of
modulo
, denoted by
To begin with, we'll use a well-known property of the order to get a bound on
Since and
, we know by Euler's Theorem that
However, we do not know that
is the least
satisfying
Nonetheless, it is a well known property of the order that
Therefore, we can conclude that
must be a positive divisor of
Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say Clearly, we must have that
Since
and powers of two will then cycle every four terms, we know that
Combining this relation with
, it follows that
Now, it is trivial to verify that In addition, we know that
Therefore, we conclude that
Hence, we must have
(Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly).
Now, since we have found , we know that
There are two good ways to finish from here:
The first way is to use a trick involving powers of Notice that
Certainly
In addition, since we already computed
, we know that
Therefore, since
and
, we conclude that
The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of into groups:
It is easy to see that
is congruent to
modulo
Now, for , notice that there are
terms in the summation, each with a different remainder upon division by
Since each of these remainders is certainly relatively prime to
, these
remainders correspond to the
positive integers less than
that are relatively prime to
Therefore,
Then, since
is divisible by
and
, it follows that
is divisible by
Therefore,
Solution 3
Obviously, will have to repeat at some point, and our goal is just to find when it repeats. Suppose
is the first time the powers of 2 repeat mod 1000, and that it is the same as
where
We have
We can factor out a
to get
Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since
has to be odd, this necessarily means
for
This means that 125 has to divide
We wish to find the minimum
such that this is true, or similarly, we just wish to find
Note that by Euler's Totient Theorem, that
After checking, and seeing that
, this directly means that
or
Since
, the smallest
pair is
What this really means is that the sequence of remainders will start repeating at
or namely that
are all distinct residues mod 1000. Adding these yields
Solution 4 (Not rigorous)
If we write out the remainders when powers of are divided by
, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down.
The answer extraction of the problem is equivalent to asking for , where
is the first number whose remainder when divided by
is a repeat. This expression is equivalent to
. So our answer will just be the first repeated remainder minus
.
(Everything up to this point has been perfectly rigorous; now, we will destroy this.)
Note that ,
and
cannot be the first repeated remainders, due to the
,
and
divisibility rules, respectively (i.e.
,
and
are the only powers of
that leave a remainder of
,
and
, respectively, when divided by
.) There is no reason why
could not be the first repeated remainder, so it probably is. Thus, our answer is
. (In fact, one can quite easily show that if
reappears at all in the sequence, it must be the first repeated remainder.)
~ihatemath123
Solution 5 (Very bad idea)
We can simply list out the last digits of all the powers of
until we find a pattern.
Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way
We can see that after comes
which can be written as
. That means that all the terms after
will repeat but negated modulo
.
Note that
. As we are looking for the sum of all of the possible values mod
, the answer is just
~EvanZ
Solution 6 (not rigorous)
Note that the problem essentially asks for the sum of all in the residue class of
that come from a power of
. We can split up this residue class into finding a solution for
in the modular system
Clearly, if
, we get the first equation to be
If
, we get our residues to be
,
, and
which add to
. Now, we just need to worry about the
mod equation. If we take a look at the powers of
modulo
, we see that we get every residue except for
. If we consider it modulo
, we see that we get every residue except for
(ie. all multiples of
). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of
in modulo
. Therefore, we can add all multiples of
from
to
and subtract the multiples of
to get
. Therefore, our final answer is
.
~Vedoral
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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