Why can you bring any number back to $9$ by doing this formula?
$begingroup$
$a$ - Digits of $a$ = $9$
If you take any number add the digits then subtract the sum of the digits from the original number and repeat until you come to a one digit number then the one digit number is always 9, for example $125$ and you add the digits $1 + 2 + 5 = $8 then do $125 - 8 = 117$ then $1 + 1 + 7 = 9$, you can always bring it back to $9$.
From what I have found, you can do this with any number that has more than one digit. But why does this happen to every number? I can't seem to find a reason.
elementary-number-theory
asked Dec 3 '18 at 19:43
LonnieLonnie
112
$endgroup$
|
show 1 more comment
$begingroup$
$a$ - Digits of $a$ = $9$
If you take any number add the digits then subtract the sum of the digits from the original number and repeat until you come to a one digit number then the one digit number is always 9, for example $125$ and you add the digits $1 + 2 + 5 = $8 then do $125 - 8 = 117$ then $1 + 1 + 7 = 9$, you can always bring it back to $9$.
From what I have found, you can do this with any number that has more than one digit. But why does this happen to every number? I can't seem to find a reason.
elementary-number-theory
asked Dec 3 '18 at 19:43
LonnieLonnie
112
$endgroup$
3
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
2
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
2
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16
|
show 1 more comment
$begingroup$
$a$ - Digits of $a$ = $9$
If you take any number add the digits then subtract the sum of the digits from the original number and repeat until you come to a one digit number then the one digit number is always 9, for example $125$ and you add the digits $1 + 2 + 5 = $8 then do $125 - 8 = 117$ then $1 + 1 + 7 = 9$, you can always bring it back to $9$.
From what I have found, you can do this with any number that has more than one digit. But why does this happen to every number? I can't seem to find a reason.
elementary-number-theory
asked Dec 3 '18 at 19:43
LonnieLonnie
112
$endgroup$
$a$ - Digits of $a$ = $9$
If you take any number add the digits then subtract the sum of the digits from the original number and repeat until you come to a one digit number then the one digit number is always 9, for example $125$ and you add the digits $1 + 2 + 5 = $8 then do $125 - 8 = 117$ then $1 + 1 + 7 = 9$, you can always bring it back to $9$.
From what I have found, you can do this with any number that has more than one digit. But why does this happen to every number? I can't seem to find a reason.
elementary-number-theory
elementary-number-theory
asked Dec 3 '18 at 19:43
LonnieLonnie
112
asked Dec 3 '18 at 19:43
LonnieLonnie
112
edited Dec 5 '18 at 18:04
Lonnie
asked Dec 3 '18 at 19:43
LonnieLonnie
112
asked Dec 3 '18 at 19:43
LonnieLonnie
112
asked Dec 3 '18 at 19:43
LonnieLonnie
112
112
3
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
2
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
2
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16
|
show 1 more comment
3
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
2
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
2
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16
3
3
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
2
2
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
2
2
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $9$ and as a result it's sum of digits is a multiple of $9$.
You do not necessarily get $9$ but you get a multiple of $9$
When repeated we will end up with nine eventually.
Same argument works for four or more digit numbers as well.
For example $$54321-15=54306$$ which is a multiple of $9$
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
add a comment |
$begingroup$
If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0leq V(j)in Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10leq xin Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_ncdot 10^n+...+x_0cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_ncdot 9V(n)+...+x_0cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))quad bullet$$ Now if $xgeq 10$ then (i) $n geq 1$ and $x_ngeq 1$ and $V(n)geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0leq j< n,$ so by $bullet $ we have $$f(x)geq (9)(x_nV(n))geq 9quad star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $ygeq 10$. So by $star$ we have $f(y)geq 9.$ That is, $10>f(y)geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $9$ and as a result it's sum of digits is a multiple of $9$.
You do not necessarily get $9$ but you get a multiple of $9$
When repeated we will end up with nine eventually.
Same argument works for four or more digit numbers as well.
For example $$54321-15=54306$$ which is a multiple of $9$
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
add a comment |
$begingroup$
Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $9$ and as a result it's sum of digits is a multiple of $9$.
You do not necessarily get $9$ but you get a multiple of $9$
When repeated we will end up with nine eventually.
Same argument works for four or more digit numbers as well.
For example $$54321-15=54306$$ which is a multiple of $9$
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
add a comment |
$begingroup$
Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $9$ and as a result it's sum of digits is a multiple of $9$.
You do not necessarily get $9$ but you get a multiple of $9$
When repeated we will end up with nine eventually.
Same argument works for four or more digit numbers as well.
For example $$54321-15=54306$$ which is a multiple of $9$
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
Note that $$100a+10b+c -(a+b+c)= 99a+9b$$ which is a multiple of $9$ and as a result it's sum of digits is a multiple of $9$.
You do not necessarily get $9$ but you get a multiple of $9$
When repeated we will end up with nine eventually.
Same argument works for four or more digit numbers as well.
For example $$54321-15=54306$$ which is a multiple of $9$
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
edited Dec 5 '18 at 22:34
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 3 '18 at 20:02
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
add a comment |
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
You may not initially get nine, but if you repeat the procedure enough you will eventually get nine.
$endgroup$
– Mike Earnest
Dec 3 '18 at 20:35
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
$begingroup$
Very good comment. Yes, that is true because you are subtracting a positive integer every time so eventually you will end up with a nine.
$endgroup$
– Mohammad Riazi-Kermani
Dec 3 '18 at 21:43
add a comment |
$begingroup$
If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0leq V(j)in Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10leq xin Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_ncdot 10^n+...+x_0cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_ncdot 9V(n)+...+x_0cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))quad bullet$$ Now if $xgeq 10$ then (i) $n geq 1$ and $x_ngeq 1$ and $V(n)geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0leq j< n,$ so by $bullet $ we have $$f(x)geq (9)(x_nV(n))geq 9quad star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $ygeq 10$. So by $star$ we have $f(y)geq 9.$ That is, $10>f(y)geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
$endgroup$
add a comment |
$begingroup$
If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0leq V(j)in Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10leq xin Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_ncdot 10^n+...+x_0cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_ncdot 9V(n)+...+x_0cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))quad bullet$$ Now if $xgeq 10$ then (i) $n geq 1$ and $x_ngeq 1$ and $V(n)geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0leq j< n,$ so by $bullet $ we have $$f(x)geq (9)(x_nV(n))geq 9quad star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $ygeq 10$. So by $star$ we have $f(y)geq 9.$ That is, $10>f(y)geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
$endgroup$
add a comment |
$begingroup$
If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0leq V(j)in Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10leq xin Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_ncdot 10^n+...+x_0cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_ncdot 9V(n)+...+x_0cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))quad bullet$$ Now if $xgeq 10$ then (i) $n geq 1$ and $x_ngeq 1$ and $V(n)geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0leq j< n,$ so by $bullet $ we have $$f(x)geq (9)(x_nV(n))geq 9quad star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $ygeq 10$. So by $star$ we have $f(y)geq 9.$ That is, $10>f(y)geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
$endgroup$
If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0leq V(j)in Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10leq xin Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_ncdot 10^n+...+x_0cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_ncdot 9V(n)+...+x_0cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))quad bullet$$ Now if $xgeq 10$ then (i) $n geq 1$ and $x_ngeq 1$ and $V(n)geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0leq j< n,$ so by $bullet $ we have $$f(x)geq (9)(x_nV(n))geq 9quad star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $ygeq 10$. So by $star$ we have $f(y)geq 9.$ That is, $10>f(y)geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
answered Dec 6 '18 at 10:29
DanielWainfleetDanielWainfleet
34.8k31648
34.8k31648
add a comment |
add a comment |
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3
$begingroup$
$1+1neq 7$ $quad$
$endgroup$
– lulu
Dec 3 '18 at 19:44
2
$begingroup$
I'm just guessing here, but I think you are after the fact that , for any $N$, $9$ divides the difference between $N$ and the sum of the digits of $N$. That's just because $10^k$ is always one more than a multiple of $9$. It's a very useful fact and the basis for lots of numerical puzzles.
$endgroup$
– lulu
Dec 3 '18 at 19:45
2
$begingroup$
Maybe you mean $1+1+7=9$?
$endgroup$
– Noah Schweber
Dec 3 '18 at 19:48
$begingroup$
yea it was a typo i was typing this fast after class and didnt see it until now
$endgroup$
– Lonnie
Dec 3 '18 at 19:57
$begingroup$
If you write your name in base 6 (instead of 10), you get the same thing but with 5 (instead of 9).... which suggests...
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:16