Why is it that whenever you multiply any number by nine the digits of the result always add up to nine? Example:
9 * 7 = 63; // 6 + 3 = 9
9 * 35 = 315; // 3 + 1 + 5 = 9
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There is a well-known divisibility test that a number is divisble by $9$ if and only if $9$ divides the sum of its digits. By multiplying a number with $9$, you are making it a multiple of $9$, and hence the sum of its digits must be divisible by $9$. Since you can iterate the digit adding process until you get down to a single digit number, the result must be divisible by $9$. The only two single digit numbers divisible by $9$ are $0$ and $9$, and you're never going to get $0$ (well, unless your original number was 0 :) ). Something similar works for $3$, but it is not as nice. If you multiply a number by 3 and then iterate the digit adding process until you have one digit, then you will wind up with $3,6$ or $9$. The reason is the same: a number is divisble by 3 if and only if the sum of its digits is. However this time when you get down to single digits, there is more than one alternative that you can arrive at. |
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The digit sum of a positive integer is always congruent to itself modulo $9$. Since the digit sum is always less than the original number, and your original number is a mutliple of $9$, repeating the process boils it down to $9$. Why? Write your integer as $a_0+10a_1+10^2a_2+\cdots 10^na_n\equiv 0\pmod{9},\quad n>0$ For every $0\leq k\leq n,\;\;10^ka_k\equiv a_k\pmod{9}$ therefore: $$a_0+10a_1+10^2a_2+\cdots 10^na_n\equiv a_0+a_1+a_2+\cdots a_n\equiv 0\pmod{9}$$ However, $$a_0+10a_1+10^2a_2+\cdots 10^na_n>a_0+a_1+a_2+\cdots a_n,\quad (n>0)$$ Thus, repeated descent brings us to smallest positive integer congruent to $0\pmod{9}$, which is $9$. |
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Start with $$9\cdot n= 9\sum_{k=0}^\infty a_k10^k=9(a_010^0+a_110^1+a_210^2+\cdots)=9a_010^0+9a_110^1+9a_210^2+\cdots$$ Then the decimal representation of $9c$ is $\left[c-1,10-c\right]$, if $c>0$. So $9a_k10^k=(a_k-1)10^{k+1}+(10-a_k)10^k$, if $a_k>0$. Adding up all digits of the prodcut will give $$ \sum (a_k-1)+(10-a_k)=\sum a_k-1+10-a_k=\sum 9. $$ |
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Let $p(x)$ be a polynomial, $r\in\mathbb{R}$. Define $q(x) := p(x)-p(r)$. Since $q(r) = 0$, we can write $q(x) = s(x)(x-r)$ for some polynomial $s(x)$. This yields $$ p(x)-p(r) =s(x)(x-r)\\ \iff p(x)/(x-r) = s(x) + p(r)/(x-r) $$ Now given any number $n\in\mathbb{N}$, there is a polynomial $p_n(x)$, such that $p_n(10) = n$. Hence $$ n/9 = p_n(10)/(10-1) = s_n(10) + p_n(1)/9$$ Hence 9 divides $n=p(10)$ if and only if 9 divides $p_n(1)$, which is the sum of all digits of $n$. |
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