What is special about square numbers here? [duplicate]The final state of 1000 light bulbs switched on/off by 1000 people passing byWord Problem Proof? (just for fun, help)Enigma : of Wizards, Dwarves and HatsCoin Arrangement Puzzlecreating a more complex sudoku (69x6)Determining the favored penny on a chessboardHow many different ways can I add three numbers to get a certain sum?Board game - winning strategyDifference PuzzlesCould someone come up with a formula explaining the following?How many ways to place three distinguishable tokens on the white spaces of a $4$-by-$4$ chess board?
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What is special about square numbers here? [duplicate]
The final state of 1000 light bulbs switched on/off by 1000 people passing byWord Problem Proof? (just for fun, help)Enigma : of Wizards, Dwarves and HatsCoin Arrangement Puzzlecreating a more complex sudoku (69x6)Determining the favored penny on a chessboardHow many different ways can I add three numbers to get a certain sum?Board game - winning strategyDifference PuzzlesCould someone come up with a formula explaining the following?How many ways to place three distinguishable tokens on the white spaces of a $4$-by-$4$ chess board?
$begingroup$
This question already has an answer here:
The final state of 1000 light bulbs switched on/off by 1000 people passing by
7 answers
I'm not not schooled in math. I'm 50 years old and I only have about a grade 8 level. But I do enjoy math and heard a question in the show "Growing Pains of a Teenage Genius" that interested me. So please forgive me. I do not speak "math."
The question has been posted here already, but I don't think the correct answer was given, and since I'm new, I haven't earned the points to be able to comment on that post. So I've started my own post.
The question is, if you have 1000 pennies lined up in a row, all heads up, and you turn over every second penny, then every third penny, then every fourth penny, etc. all the way until you turn over the thousandth and last penny, which ones will be heads up?
I've figured out that the answer is that the square numbers will be heads up. It is only the square numbers that will be flipped an even number of times to land them in the position they started out in. But I don't know why that is.
What is it about square numbers that they are the only ones that get flipped an even number of times through the process of flipping every 2nd, 3rd, 4th,...etc, penny?
I thought it must have something to do with factoring since the primes will only get flipped once, but the widening gap between each succession of flips is a bit complicated to visualize, and I don't know how to work that with factoring square numbers.
Is there something special about factoring square numbers that's applicable here?
How do you visualize this problem mathematically?
puzzle
asked Apr 13 at 22:22
DeeHDeeH
665
$endgroup$
marked as duplicate by user21820, Hans Lundmark, Lord Shark the Unknown, TheSimpliFire, Jyrki Lahtonen Apr 14 at 6:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
The final state of 1000 light bulbs switched on/off by 1000 people passing by
7 answers
I'm not not schooled in math. I'm 50 years old and I only have about a grade 8 level. But I do enjoy math and heard a question in the show "Growing Pains of a Teenage Genius" that interested me. So please forgive me. I do not speak "math."
The question has been posted here already, but I don't think the correct answer was given, and since I'm new, I haven't earned the points to be able to comment on that post. So I've started my own post.
The question is, if you have 1000 pennies lined up in a row, all heads up, and you turn over every second penny, then every third penny, then every fourth penny, etc. all the way until you turn over the thousandth and last penny, which ones will be heads up?
I've figured out that the answer is that the square numbers will be heads up. It is only the square numbers that will be flipped an even number of times to land them in the position they started out in. But I don't know why that is.
What is it about square numbers that they are the only ones that get flipped an even number of times through the process of flipping every 2nd, 3rd, 4th,...etc, penny?
I thought it must have something to do with factoring since the primes will only get flipped once, but the widening gap between each succession of flips is a bit complicated to visualize, and I don't know how to work that with factoring square numbers.
Is there something special about factoring square numbers that's applicable here?
How do you visualize this problem mathematically?
puzzle
asked Apr 13 at 22:22
DeeHDeeH
665
$endgroup$
marked as duplicate by user21820, Hans Lundmark, Lord Shark the Unknown, TheSimpliFire, Jyrki Lahtonen Apr 14 at 6:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You may find this of interest: math.stackexchange.com/questions/11223/…
$endgroup$
– Minus One-Twelfth
Apr 13 at 22:27
$begingroup$
Good question, good insights.
$endgroup$
– JonathanZ
Apr 14 at 4:00
add a comment |
$begingroup$
This question already has an answer here:
The final state of 1000 light bulbs switched on/off by 1000 people passing by
7 answers
I'm not not schooled in math. I'm 50 years old and I only have about a grade 8 level. But I do enjoy math and heard a question in the show "Growing Pains of a Teenage Genius" that interested me. So please forgive me. I do not speak "math."
The question has been posted here already, but I don't think the correct answer was given, and since I'm new, I haven't earned the points to be able to comment on that post. So I've started my own post.
The question is, if you have 1000 pennies lined up in a row, all heads up, and you turn over every second penny, then every third penny, then every fourth penny, etc. all the way until you turn over the thousandth and last penny, which ones will be heads up?
I've figured out that the answer is that the square numbers will be heads up. It is only the square numbers that will be flipped an even number of times to land them in the position they started out in. But I don't know why that is.
What is it about square numbers that they are the only ones that get flipped an even number of times through the process of flipping every 2nd, 3rd, 4th,...etc, penny?
I thought it must have something to do with factoring since the primes will only get flipped once, but the widening gap between each succession of flips is a bit complicated to visualize, and I don't know how to work that with factoring square numbers.
Is there something special about factoring square numbers that's applicable here?
How do you visualize this problem mathematically?
puzzle
asked Apr 13 at 22:22
DeeHDeeH
665
$endgroup$
This question already has an answer here:
The final state of 1000 light bulbs switched on/off by 1000 people passing by
7 answers
I'm not not schooled in math. I'm 50 years old and I only have about a grade 8 level. But I do enjoy math and heard a question in the show "Growing Pains of a Teenage Genius" that interested me. So please forgive me. I do not speak "math."
The question has been posted here already, but I don't think the correct answer was given, and since I'm new, I haven't earned the points to be able to comment on that post. So I've started my own post.
The question is, if you have 1000 pennies lined up in a row, all heads up, and you turn over every second penny, then every third penny, then every fourth penny, etc. all the way until you turn over the thousandth and last penny, which ones will be heads up?
I've figured out that the answer is that the square numbers will be heads up. It is only the square numbers that will be flipped an even number of times to land them in the position they started out in. But I don't know why that is.
What is it about square numbers that they are the only ones that get flipped an even number of times through the process of flipping every 2nd, 3rd, 4th,...etc, penny?
I thought it must have something to do with factoring since the primes will only get flipped once, but the widening gap between each succession of flips is a bit complicated to visualize, and I don't know how to work that with factoring square numbers.
Is there something special about factoring square numbers that's applicable here?
How do you visualize this problem mathematically?
This question already has an answer here:
The final state of 1000 light bulbs switched on/off by 1000 people passing by
7 answers
puzzle
puzzle
asked Apr 13 at 22:22
DeeHDeeH
665
asked Apr 13 at 22:22
DeeHDeeH
665
asked Apr 13 at 22:22
DeeHDeeH
665
asked Apr 13 at 22:22
DeeHDeeH
665
asked Apr 13 at 22:22
DeeHDeeH
665
665
marked as duplicate by user21820, Hans Lundmark, Lord Shark the Unknown, TheSimpliFire, Jyrki Lahtonen Apr 14 at 6:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by user21820, Hans Lundmark, Lord Shark the Unknown, TheSimpliFire, Jyrki Lahtonen Apr 14 at 6:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You may find this of interest: math.stackexchange.com/questions/11223/…
$endgroup$
– Minus One-Twelfth
Apr 13 at 22:27
$begingroup$
Good question, good insights.
$endgroup$
– JonathanZ
Apr 14 at 4:00
add a comment |
1
$begingroup$
You may find this of interest: math.stackexchange.com/questions/11223/…
$endgroup$
– Minus One-Twelfth
Apr 13 at 22:27
$begingroup$
Good question, good insights.
$endgroup$
– JonathanZ
Apr 14 at 4:00
1
1
$begingroup$
You may find this of interest: math.stackexchange.com/questions/11223/…
$endgroup$
– Minus One-Twelfth
Apr 13 at 22:27
$begingroup$
You may find this of interest: math.stackexchange.com/questions/11223/…
$endgroup$
– Minus One-Twelfth
Apr 13 at 22:27
$begingroup$
Good question, good insights.
$endgroup$
– JonathanZ
Apr 14 at 4:00
$begingroup$
Good question, good insights.
$endgroup$
– JonathanZ
Apr 14 at 4:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).
Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.
All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.
Let's look at $12$ for an example.
$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.
Now, let's look at a square number as an example like $16$.
$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
$endgroup$
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
$endgroup$
– DeeH
Apr 19 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).
Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.
All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.
Let's look at $12$ for an example.
$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.
Now, let's look at a square number as an example like $16$.
$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
$endgroup$
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
$endgroup$
– DeeH
Apr 19 at 14:48
add a comment |
$begingroup$
Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).
Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.
All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.
Let's look at $12$ for an example.
$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.
Now, let's look at a square number as an example like $16$.
$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
$endgroup$
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
$endgroup$
– DeeH
Apr 19 at 14:48
add a comment |
$begingroup$
Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).
Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.
All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.
Let's look at $12$ for an example.
$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.
Now, let's look at a square number as an example like $16$.
$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
$endgroup$
Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).
Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.
All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.
Let's look at $12$ for an example.
$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.
Now, let's look at a square number as an example like $16$.
$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
answered Apr 13 at 22:37
JMoravitzJMoravitz
50.8k44092
50.8k44092
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
$endgroup$
– DeeH
Apr 19 at 14:48
add a comment |
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
$endgroup$
– DeeH
Apr 19 at 14:48
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Thank you. You explained this incredibly simply and well. I get all of it, even the math language part.
$endgroup$
– DeeH
Apr 15 at 2:27
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
$begingroup$
Are you able to explain mathematically how it is that when following the sequence of flipping every 2nd, 3rd, 4th, etc coin that they are flipped according to either their even or odd number of divisors? I don't know how to reconcile that pattern with the result mathematically.
$endgroup$
– DeeH
Apr 15 at 14:37
1
1
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
When flipping every $d$'th coin, every position $n$ which has $d$ as a divisor will be flipped. For example, when flipping every second coin every even number position (which are those positions which have $2$ as a divisor). It follows then that the number of times that a coin is flipped is directly correlated to how many divisors it has.
$endgroup$
– JMoravitz
Apr 15 at 15:18
$begingroup$
This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
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– DeeH
Apr 19 at 14:48
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This is actually a very simple puzzle if you turn the pennies into numbers. The first penny is the number 1. The second penny is the number 2. Etc until the 1000th penny is the number 1000. It's just asking if you know the difference between factoring square numbers and other numbers, then to apply it in a word problem. The only difficult thing is to be able recognize what the question really is.
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– DeeH
Apr 19 at 14:48
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You may find this of interest: math.stackexchange.com/questions/11223/…
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– Minus One-Twelfth
Apr 13 at 22:27
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Good question, good insights.
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– JonathanZ
Apr 14 at 4:00