Mixing
A number from 1 to 1000 is selected. What is P(The Last Two Digits of the Cube = 1)?
$begingroup$
Y.A. Rozanov. Probability Theory: A Concise Course Chapter 1 Problem 9.
A number from 1 to 1000 is selected at random. What is probability that the last two digits of it's cube are equal to 1?
The book reports that the answer is .01.
I believe the answer follows from the fact that number $100x^2+10y+z$ cubed has the form $(100x^2+10y+z)^3$ = A trinomial expansion and using this knowledge to somehow show that there are only two solutions and each is 1/10 likely.
probability elementary-number-theory
asked Jan 27 '13 at 6:02
studentstudent
446
$endgroup$
|
show 4 more comments
$begingroup$
Y.A. Rozanov. Probability Theory: A Concise Course Chapter 1 Problem 9.
A number from 1 to 1000 is selected at random. What is probability that the last two digits of it's cube are equal to 1?
The book reports that the answer is .01.
I believe the answer follows from the fact that number $100x^2+10y+z$ cubed has the form $(100x^2+10y+z)^3$ = A trinomial expansion and using this knowledge to somehow show that there are only two solutions and each is 1/10 likely.
probability elementary-number-theory
asked Jan 27 '13 at 6:02
studentstudent
446
$endgroup$
$begingroup$
What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
1
$begingroup$
1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as1). 2. Remove those which can't generate11after cube.
$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
$begingroup$
@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
$endgroup$
– student
Jan 27 '13 at 6:11
$begingroup$
@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
1
$begingroup$
@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
$endgroup$
– student
Jan 27 '13 at 6:36
|
show 4 more comments
$begingroup$
Y.A. Rozanov. Probability Theory: A Concise Course Chapter 1 Problem 9.
A number from 1 to 1000 is selected at random. What is probability that the last two digits of it's cube are equal to 1?
The book reports that the answer is .01.
I believe the answer follows from the fact that number $100x^2+10y+z$ cubed has the form $(100x^2+10y+z)^3$ = A trinomial expansion and using this knowledge to somehow show that there are only two solutions and each is 1/10 likely.
probability elementary-number-theory
asked Jan 27 '13 at 6:02
studentstudent
446
$endgroup$
Y.A. Rozanov. Probability Theory: A Concise Course Chapter 1 Problem 9.
A number from 1 to 1000 is selected at random. What is probability that the last two digits of it's cube are equal to 1?
The book reports that the answer is .01.
I believe the answer follows from the fact that number $100x^2+10y+z$ cubed has the form $(100x^2+10y+z)^3$ = A trinomial expansion and using this knowledge to somehow show that there are only two solutions and each is 1/10 likely.
probability elementary-number-theory
probability elementary-number-theory
asked Jan 27 '13 at 6:02
studentstudent
446
asked Jan 27 '13 at 6:02
studentstudent
446
asked Jan 27 '13 at 6:02
studentstudent
446
asked Jan 27 '13 at 6:02
studentstudent
446
asked Jan 27 '13 at 6:02
studentstudent
446
446
$begingroup$
What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
1
$begingroup$
1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as1). 2. Remove those which can't generate11after cube.
$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
$begingroup$
@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
$endgroup$
– student
Jan 27 '13 at 6:11
$begingroup$
@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
1
$begingroup$
@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
$endgroup$
– student
Jan 27 '13 at 6:36
|
show 4 more comments
$begingroup$
What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
1
$begingroup$
1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as1). 2. Remove those which can't generate11after cube.
$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
$begingroup$
@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
$endgroup$
– student
Jan 27 '13 at 6:11
$begingroup$
@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
1
$begingroup$
@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
$endgroup$
– student
Jan 27 '13 at 6:36
$begingroup$
What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
$begingroup$
What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
1
1
$begingroup$
1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as
1). 2. Remove those which can't generate 11 after cube.$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
$begingroup$
1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as
1). 2. Remove those which can't generate 11 after cube.$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
$begingroup$
@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
$endgroup$
– student
Jan 27 '13 at 6:11
$begingroup$
@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
$endgroup$
– student
Jan 27 '13 at 6:11
$begingroup$
@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
$begingroup$
@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
1
1
$begingroup$
@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
$endgroup$
– student
Jan 27 '13 at 6:36
$begingroup$
@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
$endgroup$
– student
Jan 27 '13 at 6:36
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$begin{align*}
n^3&=(10m+c)^3\
&=1000m^3+300m^2c+30mc^2+c^3\
&=100left(10m^3+3m^2cright)+30mc^2+c^3;.
end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
add a comment |
$begingroup$
Hint: The number $3$ has no common divisors with $phi(100)=|mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones.
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
$endgroup$
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
|
show 1 more comment
Your Answer
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$begin{align*}
n^3&=(10m+c)^3\
&=1000m^3+300m^2c+30mc^2+c^3\
&=100left(10m^3+3m^2cright)+30mc^2+c^3;.
end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
add a comment |
$begingroup$
HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$begin{align*}
n^3&=(10m+c)^3\
&=1000m^3+300m^2c+30mc^2+c^3\
&=100left(10m^3+3m^2cright)+30mc^2+c^3;.
end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
add a comment |
$begingroup$
HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$begin{align*}
n^3&=(10m+c)^3\
&=1000m^3+300m^2c+30mc^2+c^3\
&=100left(10m^3+3m^2cright)+30mc^2+c^3;.
end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$begin{align*}
n^3&=(10m+c)^3\
&=1000m^3+300m^2c+30mc^2+c^3\
&=100left(10m^3+3m^2cright)+30mc^2+c^3;.
end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
answered Jan 27 '13 at 6:14
Brian M. ScottBrian M. Scott
461k40518920
461k40518920
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
add a comment |
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
b=7. So the only solution that gives the correct answer is b=7 and c=1. And each of those numbers is taken from 10 possible digits and are taken independently, so the probability is (1/10)(1/10). Thank You Again.
$endgroup$
– student
Jan 27 '13 at 6:22
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
$begingroup$
@student: You’re welcome. Yes, $b=7$ is the only possibility, so you get $10$ such numbers.
$endgroup$
– Brian M. Scott
Jan 27 '13 at 6:23
add a comment |
$begingroup$
Hint: The number $3$ has no common divisors with $phi(100)=|mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones.
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
$endgroup$
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
|
show 1 more comment
$begingroup$
Hint: The number $3$ has no common divisors with $phi(100)=|mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones.
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
$endgroup$
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
|
show 1 more comment
$begingroup$
Hint: The number $3$ has no common divisors with $phi(100)=|mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones.
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
$endgroup$
Hint: The number $3$ has no common divisors with $phi(100)=|mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones.
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
edited Jan 28 '13 at 22:11
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
answered Jan 27 '13 at 6:58
Jyrki LahtonenJyrki Lahtonen
110k13172391
110k13172391
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I don't understand. I'm not following the notation either. I am curious though.
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– student
Jan 27 '13 at 7:02
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What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
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@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
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– Jyrki Lahtonen
Jan 27 '13 at 7:06
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The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
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– Jyrki Lahtonen
Jan 27 '13 at 7:09
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I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
|
show 1 more comment
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
I don't understand. I'm not following the notation either. I am curious though.
$endgroup$
– student
Jan 27 '13 at 7:02
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
What theorem is this?
$endgroup$
– student
Jan 27 '13 at 7:04
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
@student: Consider the numbers from 1 to 100. Throw away those that are divisible by either 2 or 5. Forty numbers remain. My hint depends on a little bit of group theory. In an abelian group of order 40, raising the elements to third power will permute the elements. Meaning that exactly one of those forty numbers has a cube that ends with 11. Exactly the same thing will happen once in each century.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:06
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
The result being used is that because $gcd(3,40)=1$, the homomorphism $f:Ato A, xmapsto x^3$ from a finite abelian group $A$ to itself has a trivial kernel. Therefore the freshman theorem implies that the image of $f$ has 40 elements. Therefore it is a bijection.
$endgroup$
– Jyrki Lahtonen
Jan 27 '13 at 7:09
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
$begingroup$
I can't follow this argument. It is too advanced. Thanks. I want to study group theory some day. Is there a good book on it you could recommend?
$endgroup$
– student
Jan 27 '13 at 7:18
|
show 1 more comment
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What prevents you from carrying out your ideas? Note that you can "throw away" many of the terms in the expansion because they are divisible by 100. (You are working mod 100.)
$endgroup$
– Potato
Jan 27 '13 at 6:08
1
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1. Consider how you can get the unit digit to be one after the cube(All numbers will have unit digit as
1). 2. Remove those which can't generate11after cube.$endgroup$
– hjpotter92
Jan 27 '13 at 6:09
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@Potato I know. I think I'm just tired and frustrated. I think the only term that can contribute is the z^3 because the 30y^2z +30z^2y terms are too large. I know this should be easy.
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– student
Jan 27 '13 at 6:11
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@BackinaFlash z must be 1 I suppose. Then ....
$endgroup$
– student
Jan 27 '13 at 6:13
1
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@BackinaFlash Yes. But now I am questioning my logic. I argued that the answer is .01 because 7 and 1 can each be taken 1 way out of 10 possibilities. If you where simply guessing digits you would have a 1/100 chance of guessing 7 and 1. But now I think it would be better to have said that there are only 10 possibilities out of 1000 because there are only 10 ways to end the digit with "71". I like that argument more.
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– student
Jan 27 '13 at 6:36