Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$












19














Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
begin{align*}
360 ÷ 2 &= 180 text{, and } 1 + 8 + 0 = 9\
180 ÷ 2 &= 90 text{, and } 9 + 0 = 9\
90 ÷ 2 &= 45 text{, and } 4 + 5 = 9\
45 ÷ 2 &= 22.5 text{, and } 2 + 2 + 5 = 9\
22.5 ÷ 2 &= 11.25 text{, and } 1 + 1 + 2 + 5 = 9\
11.25 ÷ 2 &= 5.625 text{, and } 5 + 6 + 2 + 5 = 18 text{, and } 1 + 8 = 9\
5.625 ÷ 2 &= 2.8125 text{, and } 2 + 8 + 1 + 2 + 5 = 18 text{, and } 1 + 8 = 9
end{align*}



As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!










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  • 14




    One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
    – Matti P.
    yesterday






  • 8




    Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
    – Shalop
    yesterday












  • It would work too with division by $5$ (or multiplication by $2$)
    – Henry
    yesterday








  • 1




    Try dividing by 11 and see what you get :-)
    – ChatterOne
    19 hours ago
















19














Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
begin{align*}
360 ÷ 2 &= 180 text{, and } 1 + 8 + 0 = 9\
180 ÷ 2 &= 90 text{, and } 9 + 0 = 9\
90 ÷ 2 &= 45 text{, and } 4 + 5 = 9\
45 ÷ 2 &= 22.5 text{, and } 2 + 2 + 5 = 9\
22.5 ÷ 2 &= 11.25 text{, and } 1 + 1 + 2 + 5 = 9\
11.25 ÷ 2 &= 5.625 text{, and } 5 + 6 + 2 + 5 = 18 text{, and } 1 + 8 = 9\
5.625 ÷ 2 &= 2.8125 text{, and } 2 + 8 + 1 + 2 + 5 = 18 text{, and } 1 + 8 = 9
end{align*}



As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!










share|cite|improve this question




















  • 14




    One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
    – Matti P.
    yesterday






  • 8




    Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
    – Shalop
    yesterday












  • It would work too with division by $5$ (or multiplication by $2$)
    – Henry
    yesterday








  • 1




    Try dividing by 11 and see what you get :-)
    – ChatterOne
    19 hours ago














19












19








19


2





Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
begin{align*}
360 ÷ 2 &= 180 text{, and } 1 + 8 + 0 = 9\
180 ÷ 2 &= 90 text{, and } 9 + 0 = 9\
90 ÷ 2 &= 45 text{, and } 4 + 5 = 9\
45 ÷ 2 &= 22.5 text{, and } 2 + 2 + 5 = 9\
22.5 ÷ 2 &= 11.25 text{, and } 1 + 1 + 2 + 5 = 9\
11.25 ÷ 2 &= 5.625 text{, and } 5 + 6 + 2 + 5 = 18 text{, and } 1 + 8 = 9\
5.625 ÷ 2 &= 2.8125 text{, and } 2 + 8 + 1 + 2 + 5 = 18 text{, and } 1 + 8 = 9
end{align*}



As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!










share|cite|improve this question















Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
begin{align*}
360 ÷ 2 &= 180 text{, and } 1 + 8 + 0 = 9\
180 ÷ 2 &= 90 text{, and } 9 + 0 = 9\
90 ÷ 2 &= 45 text{, and } 4 + 5 = 9\
45 ÷ 2 &= 22.5 text{, and } 2 + 2 + 5 = 9\
22.5 ÷ 2 &= 11.25 text{, and } 1 + 1 + 2 + 5 = 9\
11.25 ÷ 2 &= 5.625 text{, and } 5 + 6 + 2 + 5 = 18 text{, and } 1 + 8 = 9\
5.625 ÷ 2 &= 2.8125 text{, and } 2 + 8 + 1 + 2 + 5 = 18 text{, and } 1 + 8 = 9
end{align*}



As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!







elementary-number-theory decimal-expansion






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edited 17 hours ago









Asaf Karagila

302k32427757




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asked yesterday









rosarosa

492515




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  • 14




    One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
    – Matti P.
    yesterday






  • 8




    Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
    – Shalop
    yesterday












  • It would work too with division by $5$ (or multiplication by $2$)
    – Henry
    yesterday








  • 1




    Try dividing by 11 and see what you get :-)
    – ChatterOne
    19 hours ago














  • 14




    One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
    – Matti P.
    yesterday






  • 8




    Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
    – Shalop
    yesterday












  • It would work too with division by $5$ (or multiplication by $2$)
    – Henry
    yesterday








  • 1




    Try dividing by 11 and see what you get :-)
    – ChatterOne
    19 hours ago








14




14




One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
– Matti P.
yesterday




One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/…
– Matti P.
yesterday




8




8




Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
– Shalop
yesterday






Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9.
– Shalop
yesterday














It would work too with division by $5$ (or multiplication by $2$)
– Henry
yesterday






It would work too with division by $5$ (or multiplication by $2$)
– Henry
yesterday






1




1




Try dividing by 11 and see what you get :-)
– ChatterOne
19 hours ago




Try dividing by 11 and see what you get :-)
– ChatterOne
19 hours ago










6 Answers
6






active

oldest

votes


















27














Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
$$
n=17254,quad
d(n)=1+7+2+5+4=19,
quad
d(d(n))=1+9=10,
quad
d(d(d(n)))=1+0=1=d^*(n)
$$

Note that $d(n)le n$, equality holding if and only if $1le nle 9$. Also, $1le d^*(n)le 9$, because the process stops only when the digit sum obtained is a one-digit number.



The main point is that $n-d(n)$ is divisible by $9$: indeed, if
$$
n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
$$

then
$$
n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+dots+a_n(10^n-1)
$$

and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
$$
n-d^*(n)=bigl(n-d(n)bigr)+bigl(d(n)-d(d(n))bigr)+bigl(d(d(n))-d(d(d(n)))bigr)
$$

and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.



Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that




$d^*(n)=9$ if and only if $n$ is divisible by $9$.




Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.



When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.



What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
begin{align}
&45 xrightarrow{cdot10} 450 xrightarrow{/2} 225 && d^*(225)=9 \
&225 xrightarrow{cdot10} 2250 xrightarrow{/2} 1225 && d^*(1225)=9
end{align}

and so on. Note that the first step can also be stated as
$$
45 xrightarrow{cdot5} 225
$$



Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.






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    15














    In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.



    What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.



    A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.



    More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.






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    • 2




      I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
      – Paŭlo Ebermann
      yesterday






    • 4




      @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
      – Adam Bailey
      yesterday



















    7














    A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.






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      6














      I think the other answers are unnecessarily complicated.



      It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.



      And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.






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      • But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
        – Adam Bailey
        17 hours ago










      • @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
        – Pedro A
        14 hours ago










      • @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
        – Adam Bailey
        13 hours ago










      • @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
        – Pedro A
        13 hours ago










      • @PedroA Ah, I understand now.
        – Adam Bailey
        11 hours ago



















      1














      1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.



      Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.



      If $N = sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.



      And $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$



      $(sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$



      $sum_{k=0}^n [a_k*10^k - underbrace{999...9}k*a_k]=$



      $sum_{k=0}^n a_k(10^k - underbrace{999...9}k)=sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.



      2) we can extend that decimals:



      If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.



      Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.



      3) If $N$ is a $9$ times a terminating decimal than $frac N2$ is $9$ times a terminating decimal.



      If $N = 9times M$ and $M$ is a terminating decimal then $frac M2$ is a terminating decimal. Then $frac N2 = 9times frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.



      4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.






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        1














        Just a quick addendum and question about the division by 7.
        One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.



        For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.



        This suggests an extension along this line.






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        • This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
          – rosa
          2 hours ago











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        6 Answers
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        27














        Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
        $$
        n=17254,quad
        d(n)=1+7+2+5+4=19,
        quad
        d(d(n))=1+9=10,
        quad
        d(d(d(n)))=1+0=1=d^*(n)
        $$

        Note that $d(n)le n$, equality holding if and only if $1le nle 9$. Also, $1le d^*(n)le 9$, because the process stops only when the digit sum obtained is a one-digit number.



        The main point is that $n-d(n)$ is divisible by $9$: indeed, if
        $$
        n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
        $$

        then
        $$
        n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+dots+a_n(10^n-1)
        $$

        and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
        $$
        n-d^*(n)=bigl(n-d(n)bigr)+bigl(d(n)-d(d(n))bigr)+bigl(d(d(n))-d(d(d(n)))bigr)
        $$

        and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.



        Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that




        $d^*(n)=9$ if and only if $n$ is divisible by $9$.




        Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.



        When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.



        What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
        begin{align}
        &45 xrightarrow{cdot10} 450 xrightarrow{/2} 225 && d^*(225)=9 \
        &225 xrightarrow{cdot10} 2250 xrightarrow{/2} 1225 && d^*(1225)=9
        end{align}

        and so on. Note that the first step can also be stated as
        $$
        45 xrightarrow{cdot5} 225
        $$



        Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.






        share|cite|improve this answer


























          27














          Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
          $$
          n=17254,quad
          d(n)=1+7+2+5+4=19,
          quad
          d(d(n))=1+9=10,
          quad
          d(d(d(n)))=1+0=1=d^*(n)
          $$

          Note that $d(n)le n$, equality holding if and only if $1le nle 9$. Also, $1le d^*(n)le 9$, because the process stops only when the digit sum obtained is a one-digit number.



          The main point is that $n-d(n)$ is divisible by $9$: indeed, if
          $$
          n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
          $$

          then
          $$
          n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+dots+a_n(10^n-1)
          $$

          and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
          $$
          n-d^*(n)=bigl(n-d(n)bigr)+bigl(d(n)-d(d(n))bigr)+bigl(d(d(n))-d(d(d(n)))bigr)
          $$

          and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.



          Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that




          $d^*(n)=9$ if and only if $n$ is divisible by $9$.




          Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.



          When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.



          What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
          begin{align}
          &45 xrightarrow{cdot10} 450 xrightarrow{/2} 225 && d^*(225)=9 \
          &225 xrightarrow{cdot10} 2250 xrightarrow{/2} 1225 && d^*(1225)=9
          end{align}

          and so on. Note that the first step can also be stated as
          $$
          45 xrightarrow{cdot5} 225
          $$



          Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.






          share|cite|improve this answer
























            27












            27








            27






            Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
            $$
            n=17254,quad
            d(n)=1+7+2+5+4=19,
            quad
            d(d(n))=1+9=10,
            quad
            d(d(d(n)))=1+0=1=d^*(n)
            $$

            Note that $d(n)le n$, equality holding if and only if $1le nle 9$. Also, $1le d^*(n)le 9$, because the process stops only when the digit sum obtained is a one-digit number.



            The main point is that $n-d(n)$ is divisible by $9$: indeed, if
            $$
            n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
            $$

            then
            $$
            n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+dots+a_n(10^n-1)
            $$

            and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
            $$
            n-d^*(n)=bigl(n-d(n)bigr)+bigl(d(n)-d(d(n))bigr)+bigl(d(d(n))-d(d(d(n)))bigr)
            $$

            and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.



            Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that




            $d^*(n)=9$ if and only if $n$ is divisible by $9$.




            Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.



            When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.



            What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
            begin{align}
            &45 xrightarrow{cdot10} 450 xrightarrow{/2} 225 && d^*(225)=9 \
            &225 xrightarrow{cdot10} 2250 xrightarrow{/2} 1225 && d^*(1225)=9
            end{align}

            and so on. Note that the first step can also be stated as
            $$
            45 xrightarrow{cdot5} 225
            $$



            Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.






            share|cite|improve this answer












            Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
            $$
            n=17254,quad
            d(n)=1+7+2+5+4=19,
            quad
            d(d(n))=1+9=10,
            quad
            d(d(d(n)))=1+0=1=d^*(n)
            $$

            Note that $d(n)le n$, equality holding if and only if $1le nle 9$. Also, $1le d^*(n)le 9$, because the process stops only when the digit sum obtained is a one-digit number.



            The main point is that $n-d(n)$ is divisible by $9$: indeed, if
            $$
            n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
            $$

            then
            $$
            n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+dots+a_n(10^n-1)
            $$

            and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above)
            $$
            n-d^*(n)=bigl(n-d(n)bigr)+bigl(d(n)-d(d(n))bigr)+bigl(d(d(n))-d(d(d(n)))bigr)
            $$

            and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.



            Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that




            $d^*(n)=9$ if and only if $n$ is divisible by $9$.




            Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.



            When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.



            What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually
            begin{align}
            &45 xrightarrow{cdot10} 450 xrightarrow{/2} 225 && d^*(225)=9 \
            &225 xrightarrow{cdot10} 2250 xrightarrow{/2} 1225 && d^*(1225)=9
            end{align}

            and so on. Note that the first step can also be stated as
            $$
            45 xrightarrow{cdot5} 225
            $$



            Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            egregegreg

            179k1485202




            179k1485202























                15














                In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.



                What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.



                A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.



                More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.






                share|cite|improve this answer



















                • 2




                  I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                  – Paŭlo Ebermann
                  yesterday






                • 4




                  @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                  – Adam Bailey
                  yesterday
















                15














                In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.



                What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.



                A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.



                More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.






                share|cite|improve this answer



















                • 2




                  I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                  – Paŭlo Ebermann
                  yesterday






                • 4




                  @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                  – Adam Bailey
                  yesterday














                15












                15








                15






                In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.



                What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.



                A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.



                More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.






                share|cite|improve this answer














                In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.



                What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.



                A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.



                More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 17 hours ago

























                answered yesterday









                Adam BaileyAdam Bailey

                2,0721319




                2,0721319








                • 2




                  I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                  – Paŭlo Ebermann
                  yesterday






                • 4




                  @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                  – Adam Bailey
                  yesterday














                • 2




                  I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                  – Paŭlo Ebermann
                  yesterday






                • 4




                  @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                  – Adam Bailey
                  yesterday








                2




                2




                I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                – Paŭlo Ebermann
                yesterday




                I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits.
                – Paŭlo Ebermann
                yesterday




                4




                4




                @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                – Adam Bailey
                yesterday




                @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined?
                – Adam Bailey
                yesterday











                7














                A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.






                share|cite|improve this answer


























                  7














                  A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.






                  share|cite|improve this answer
























                    7












                    7








                    7






                    A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.






                    share|cite|improve this answer












                    A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    ncmathsadistncmathsadist

                    42.4k259102




                    42.4k259102























                        6














                        I think the other answers are unnecessarily complicated.



                        It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.



                        And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.






                        share|cite|improve this answer























                        • But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                          – Adam Bailey
                          17 hours ago










                        • @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                          – Pedro A
                          14 hours ago










                        • @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                          – Adam Bailey
                          13 hours ago










                        • @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                          – Pedro A
                          13 hours ago










                        • @PedroA Ah, I understand now.
                          – Adam Bailey
                          11 hours ago
















                        6














                        I think the other answers are unnecessarily complicated.



                        It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.



                        And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.






                        share|cite|improve this answer























                        • But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                          – Adam Bailey
                          17 hours ago










                        • @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                          – Pedro A
                          14 hours ago










                        • @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                          – Adam Bailey
                          13 hours ago










                        • @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                          – Pedro A
                          13 hours ago










                        • @PedroA Ah, I understand now.
                          – Adam Bailey
                          11 hours ago














                        6












                        6








                        6






                        I think the other answers are unnecessarily complicated.



                        It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.



                        And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.






                        share|cite|improve this answer














                        I think the other answers are unnecessarily complicated.



                        It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.



                        And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited yesterday

























                        answered yesterday









                        Pedro APedro A

                        2,0161826




                        2,0161826












                        • But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                          – Adam Bailey
                          17 hours ago










                        • @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                          – Pedro A
                          14 hours ago










                        • @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                          – Adam Bailey
                          13 hours ago










                        • @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                          – Pedro A
                          13 hours ago










                        • @PedroA Ah, I understand now.
                          – Adam Bailey
                          11 hours ago


















                        • But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                          – Adam Bailey
                          17 hours ago










                        • @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                          – Pedro A
                          14 hours ago










                        • @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                          – Adam Bailey
                          13 hours ago










                        • @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                          – Pedro A
                          13 hours ago










                        • @PedroA Ah, I understand now.
                          – Adam Bailey
                          11 hours ago
















                        But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                        – Adam Bailey
                        17 hours ago




                        But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal.
                        – Adam Bailey
                        17 hours ago












                        @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                        – Pedro A
                        14 hours ago




                        @AdamBailey Well, I agree. But what is your point? Why are you considering 7?
                        – Pedro A
                        14 hours ago












                        @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                        – Adam Bailey
                        13 hours ago




                        @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9.
                        – Adam Bailey
                        13 hours ago












                        @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                        – Pedro A
                        13 hours ago




                        @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7.
                        – Pedro A
                        13 hours ago












                        @PedroA Ah, I understand now.
                        – Adam Bailey
                        11 hours ago




                        @PedroA Ah, I understand now.
                        – Adam Bailey
                        11 hours ago











                        1














                        1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.



                        Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.



                        If $N = sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.



                        And $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$



                        $(sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$



                        $sum_{k=0}^n [a_k*10^k - underbrace{999...9}k*a_k]=$



                        $sum_{k=0}^n a_k(10^k - underbrace{999...9}k)=sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.



                        2) we can extend that decimals:



                        If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.



                        Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.



                        3) If $N$ is a $9$ times a terminating decimal than $frac N2$ is $9$ times a terminating decimal.



                        If $N = 9times M$ and $M$ is a terminating decimal then $frac M2$ is a terminating decimal. Then $frac N2 = 9times frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.



                        4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.






                        share|cite|improve this answer


























                          1














                          1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.



                          Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.



                          If $N = sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.



                          And $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$



                          $(sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$



                          $sum_{k=0}^n [a_k*10^k - underbrace{999...9}k*a_k]=$



                          $sum_{k=0}^n a_k(10^k - underbrace{999...9}k)=sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.



                          2) we can extend that decimals:



                          If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.



                          Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.



                          3) If $N$ is a $9$ times a terminating decimal than $frac N2$ is $9$ times a terminating decimal.



                          If $N = 9times M$ and $M$ is a terminating decimal then $frac M2$ is a terminating decimal. Then $frac N2 = 9times frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.



                          4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.



                            Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.



                            If $N = sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.



                            And $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$



                            $(sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$



                            $sum_{k=0}^n [a_k*10^k - underbrace{999...9}k*a_k]=$



                            $sum_{k=0}^n a_k(10^k - underbrace{999...9}k)=sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.



                            2) we can extend that decimals:



                            If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.



                            Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.



                            3) If $N$ is a $9$ times a terminating decimal than $frac N2$ is $9$ times a terminating decimal.



                            If $N = 9times M$ and $M$ is a terminating decimal then $frac M2$ is a terminating decimal. Then $frac N2 = 9times frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.



                            4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.






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                            1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.



                            Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.



                            If $N = sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.



                            And $(sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$



                            $(sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$



                            $sum_{k=0}^n [a_k*10^k - underbrace{999...9}k*a_k]=$



                            $sum_{k=0}^n a_k(10^k - underbrace{999...9}k)=sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.



                            2) we can extend that decimals:



                            If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.



                            Why? Well, its the same as above except $N$ has be divided by a power of $10$. But that only affects the position of the decimal point. It doesn't affect the digits.



                            3) If $N$ is a $9$ times a terminating decimal than $frac N2$ is $9$ times a terminating decimal.



                            If $N = 9times M$ and $M$ is a terminating decimal then $frac M2$ is a terminating decimal. Then $frac N2 = 9times frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.



                            4) Since $360 = 9*40$ we start with a number whose digits add to $9$. Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            fleabloodfleablood

                            68.6k22685




                            68.6k22685























                                1














                                Just a quick addendum and question about the division by 7.
                                One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.



                                For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.



                                This suggests an extension along this line.






                                share|cite|improve this answer





















                                • This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                  – rosa
                                  2 hours ago
















                                1














                                Just a quick addendum and question about the division by 7.
                                One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.



                                For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.



                                This suggests an extension along this line.






                                share|cite|improve this answer





















                                • This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                  – rosa
                                  2 hours ago














                                1












                                1








                                1






                                Just a quick addendum and question about the division by 7.
                                One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.



                                For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.



                                This suggests an extension along this line.






                                share|cite|improve this answer












                                Just a quick addendum and question about the division by 7.
                                One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.



                                For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.



                                This suggests an extension along this line.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 3 hours ago









                                Dr Mike EckerDr Mike Ecker

                                411




                                411












                                • This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                  – rosa
                                  2 hours ago


















                                • This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                  – rosa
                                  2 hours ago
















                                This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                – rosa
                                2 hours ago




                                This serves as a great idea for another pattern to ponder on..hope i can learn more from this...
                                – rosa
                                2 hours ago


















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