Mixing
How is the missing digit calculated?
Recently I watched a video by Arthur Benjamin:
https://youtu.be/e4PTvXtz4GM?t=337
I was curious how we solved this part of his show, and would like to know.
Essentially at this point of the video, he asks three audience members to take the number $576$ and multiply it by a $4$ digit number. Thus the resulting product is a $6$ or $7$ digit number.
Then he asks each of the members to call out all 5 of their 6, or 6 of their 7 digits, and he will find the missing digit.
The first person calls:
$8,0,9,3,8$, and Arthur guesses $8$ as the digit he leaves out.
The second person calls:
$4,7,2,5,8,4$, and Arthur guesses $6$.
The third person calls:
$9,4,4,5,4,4$, and Arthur guesses $6$.
My question is, how exactly he knew this.
Firstly, I realize that adding each of the digits together in the product yields $36$, regardless of it being a $6$ or $7$ digit number. So then I assume that he just added all their digits and subtracted from $36$ to get their missing digit.
If this is the case, where did the number $36$ come from?
I don't think this is fully true, as something like $231times 4412 = 1019172$, which has digits sum up to $21$.
I don't have any previous experience is number theory (if I need to know this to understand why it works)
number-theory elementary-number-theory
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
add a comment |
Recently I watched a video by Arthur Benjamin:
https://youtu.be/e4PTvXtz4GM?t=337
I was curious how we solved this part of his show, and would like to know.
Essentially at this point of the video, he asks three audience members to take the number $576$ and multiply it by a $4$ digit number. Thus the resulting product is a $6$ or $7$ digit number.
Then he asks each of the members to call out all 5 of their 6, or 6 of their 7 digits, and he will find the missing digit.
The first person calls:
$8,0,9,3,8$, and Arthur guesses $8$ as the digit he leaves out.
The second person calls:
$4,7,2,5,8,4$, and Arthur guesses $6$.
The third person calls:
$9,4,4,5,4,4$, and Arthur guesses $6$.
My question is, how exactly he knew this.
Firstly, I realize that adding each of the digits together in the product yields $36$, regardless of it being a $6$ or $7$ digit number. So then I assume that he just added all their digits and subtracted from $36$ to get their missing digit.
If this is the case, where did the number $36$ come from?
I don't think this is fully true, as something like $231times 4412 = 1019172$, which has digits sum up to $21$.
I don't have any previous experience is number theory (if I need to know this to understand why it works)
number-theory elementary-number-theory
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
add a comment |
Recently I watched a video by Arthur Benjamin:
https://youtu.be/e4PTvXtz4GM?t=337
I was curious how we solved this part of his show, and would like to know.
Essentially at this point of the video, he asks three audience members to take the number $576$ and multiply it by a $4$ digit number. Thus the resulting product is a $6$ or $7$ digit number.
Then he asks each of the members to call out all 5 of their 6, or 6 of their 7 digits, and he will find the missing digit.
The first person calls:
$8,0,9,3,8$, and Arthur guesses $8$ as the digit he leaves out.
The second person calls:
$4,7,2,5,8,4$, and Arthur guesses $6$.
The third person calls:
$9,4,4,5,4,4$, and Arthur guesses $6$.
My question is, how exactly he knew this.
Firstly, I realize that adding each of the digits together in the product yields $36$, regardless of it being a $6$ or $7$ digit number. So then I assume that he just added all their digits and subtracted from $36$ to get their missing digit.
If this is the case, where did the number $36$ come from?
I don't think this is fully true, as something like $231times 4412 = 1019172$, which has digits sum up to $21$.
I don't have any previous experience is number theory (if I need to know this to understand why it works)
number-theory elementary-number-theory
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
Recently I watched a video by Arthur Benjamin:
https://youtu.be/e4PTvXtz4GM?t=337
I was curious how we solved this part of his show, and would like to know.
Essentially at this point of the video, he asks three audience members to take the number $576$ and multiply it by a $4$ digit number. Thus the resulting product is a $6$ or $7$ digit number.
Then he asks each of the members to call out all 5 of their 6, or 6 of their 7 digits, and he will find the missing digit.
The first person calls:
$8,0,9,3,8$, and Arthur guesses $8$ as the digit he leaves out.
The second person calls:
$4,7,2,5,8,4$, and Arthur guesses $6$.
The third person calls:
$9,4,4,5,4,4$, and Arthur guesses $6$.
My question is, how exactly he knew this.
Firstly, I realize that adding each of the digits together in the product yields $36$, regardless of it being a $6$ or $7$ digit number. So then I assume that he just added all their digits and subtracted from $36$ to get their missing digit.
If this is the case, where did the number $36$ come from?
I don't think this is fully true, as something like $231times 4412 = 1019172$, which has digits sum up to $21$.
I don't have any previous experience is number theory (if I need to know this to understand why it works)
number-theory elementary-number-theory
number-theory elementary-number-theory
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
asked Jan 1 at 0:12
K Split XK Split X
4,21111031
4,21111031
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = left(9 + 1right)^n = 9^n + n times 9^{n - 1} + ldots + n times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.
As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.
Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
add a comment |
The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.
There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = left(9 + 1right)^n = 9^n + n times 9^{n - 1} + ldots + n times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.
As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.
Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
add a comment |
It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = left(9 + 1right)^n = 9^n + n times 9^{n - 1} + ldots + n times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.
As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.
Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
add a comment |
It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = left(9 + 1right)^n = 9^n + n times 9^{n - 1} + ldots + n times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.
As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.
Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = left(9 + 1right)^n = 9^n + n times 9^{n - 1} + ldots + n times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.
As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.
Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
edited Jan 1 at 5:33
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
answered Jan 1 at 0:15
John OmielanJohn Omielan
1,14118
1,14118
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
add a comment |
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
Good answer, +1. This shows how you don't even need it to be a four digit number that you multiply the $576$ by (although keeping the number of digits small makes it easier), and there is nothing that special about $576$ to start.
– Dave
Jan 1 at 0:29
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
@Dave Yes, that is true. However, I think the relatively small number of digits, plus possibly the value of $576$ itself, were picked so he could uniquely handle the issue of a $0$ or $9$, as Mark Bennet discusses in his answer.
– John Omielan
Jan 1 at 0:32
add a comment |
The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.
There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
add a comment |
The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.
There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
add a comment |
The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.
There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.
There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
answered Jan 1 at 0:20
Mark BennetMark Bennet
80.6k981179
80.6k981179
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
add a comment |
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
Art Benjamin has said that his strategy when the answer is $9$ or $0$ is to just guess. A fifty-fifty shot isn't bad, and even if he only gets it on the second try it's still impressive!
– Carmeister
Jan 1 at 8:21
add a comment |
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