3

Let $s(x)$ be an abstraction for casting out nines of integer $x$. For all integers $x$, prove $s(x) \equiv x$ mod $9$

I'm not asking for an answer more of a way to attack this problem. Can't think of where to start

Martin Sleziak
  • 53,687
tausch86
  • 137

1 Answers1

4

Hint: Because $10\equiv 1\pmod{9}$, we have $10^k\equiv 1\pmod 9$ for any non-negative integer $k$.

And note for example that $2013=2\times 10^3+0\times 10^2+1\times 10^1+3\times 10^0$.

Added: Consider for example the $4$-digit number "$N=a_3a_2a_1a_0$" where the $a_i$ are digits. For example, with the $4$-digit number $7145$ we have $a_0=5$, $a_1=4$, $a_2=1$ and $a_3=7$.

We want to show that $N$ and the sum $a_3+a_2+a_1+a_0$ have the same remainder on division by $9$, that is, are congruent to each other modulo $9$. Note that $$N=a_3\times 10^3+a_2\times 10^2+a_1\times 10^1+a_0\times 10^0.$$ Because $10^k\equiv 1\pmod 9$, we have $$N\equiv a_3\times 1+a_2\times 1+a_1\times 1+a_0\times 1=a_3+a_2+a_1+a_0\pmod{9},$$ which is what we wanted to show.

We can also do it without congruences. For $$(1000a_3+100a_2+10a_1+a_0)-(a_3+a_2+a_1+a_0)=a_3(999)+a_2(99)+a_1(9),$$ and the right-hand side is clearly divisible by $9$.

André Nicolas
  • 507,029
  • 1
    hm. I think I'm close to understanding. because $10^k \equiv \ 1 \ mod \ 9$ would that be an identity? $1 \times $ it's coefficient? Then we could modulo the coeffient then add up the modulus of the coefficients? I apologize if I'm not making sense either. – tausch86 Jun 20 '13 at 02:54
  • 2
    Let us do $842$. This is $8\times 10^2+4\times 10^1+2\times 10^0$, so modulo $9$ it is congruent to $8\times 1+4\times 1+2\times 1$. If you are unfomfortable with congruences, I can do another answer that doesn't mention them. – André Nicolas Jun 20 '13 at 03:02
  • 1
    I understand what you mean by congruences so far. How could I generally prove that casting out nines of a number is congruent to that number modulus nine? I like where you're headed but I don't really see the entrance to get inside. – tausch86 Jun 20 '13 at 03:12
  • 1
    I will add a bit to the answer. – André Nicolas Jun 20 '13 at 03:13