Why can you bring any number back to $9$ by doing this formula?












1












$begingroup$


$a$ - 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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


$a$ - 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







share|cite|improve this question











$endgroup$








  • 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














1












1








1


0



$begingroup$


$a$ - 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







share|cite|improve this question











$endgroup$




$a$ - 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 18:04







Lonnie

















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














  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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 $9$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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 $f$.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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 $9$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    3












    $begingroup$

    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 $9$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    3












    3








    3





    $begingroup$

    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 $9$






    share|cite|improve this answer











    $endgroup$



    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 $9$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 5 '18 at 22:34

























    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


















    • $begingroup$
      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











    0












    $begingroup$

    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 $f$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $f$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $f$.






        share|cite|improve this answer









        $endgroup$



        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 $f$.







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        answered Dec 6 '18 at 10:29









        DanielWainfleetDanielWainfleet

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