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If $N= (\underbrace{323232 \cdots}_{50 \text{ digits}})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?

I am looking for a "fast" approach that could be used to solve this in less than $2$ minutes.

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Quixotic
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    This is computing $N\mod 8$. Since the $9\mod 8 \equiv 1$, this is just the sum of digits of the number $N$, i.e. $(3+2)\times 25\mod 8 \equiv (3+2)\times 1\mod 8 = 5$. Did I miss the question ? – Sasha Aug 26 '11 at 16:45
  • @Sasha:No,you didn't miss the question.I just can't see this solution before :/ – Quixotic Aug 26 '11 at 17:26

2 Answers2

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The same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.

Why?

Because remember that writing, say, $(38571)_9$ "really" means $$1 + 7\times 9 + 5\times 9^2 + 8\times 9^3 + 3\times 9^4.$$ When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ by $8$, the remainder is $1^2 = 1$; then you divide $9^3$ by $8$, the remainder is $1^3=1$, etc. So the remainder of this number when divided by $8$ is the same as the remainder of $$1 + 7\times 1 + 5\times 9^2 + 8\times 1^3 + 3\times 1^4$$ which is the same as the remainder of $1+7+5+8+3 = 24$, which is $0$.

In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.

(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).

(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).

Arturo Magidin
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  • I feel like stupid now,the first general result is not new to me and could be easily be shown as follows:$N = (a_1a_2\cdots a_k)_r = a_k + \cdots +a_2 \times r^{k-2} + a_2 \times r^{k-1}$ and $ S = a_1 + a_2 + \cdots a_k$ then it could be easily shown that $\frac{N-S}{r-1} = I \text{ (an integer) }+\frac{S}{r-1}$,which proves that result. But I don't know why I can't see this direct application before! :/ – Quixotic Aug 26 '11 at 17:23
  • Anyways,the more general result is new to me,and perhaps a very useful one,could you hint me a easy proof for the same? – Quixotic Aug 26 '11 at 17:25
  • @FoolForMath The key idea is to learn modular arithmetic and, more generally, modular reductions of problems in quotient algebras. It's one way to algebraically divide and conquer. You can find further discussion of such in some of my posts here if you follow the link I gave, and its links... – Bill Dubuque Aug 26 '11 at 17:28
  • I understood it now :D and it was before Bill Dubuque's post :-) but yes Bill you are right,my understanding is due to modular arithmetic. – Quixotic Aug 26 '11 at 17:31
  • @Foo So you knew modular arithmetic before posting, but it was not clear how to apply it here? – Bill Dubuque Aug 26 '11 at 17:35
  • @Bill Dubuque:Yes I knew modular arithmetic before I made this post and the first general result too(from Higher algebra classes),but may be due of lack of sufficient practice of this theory,I didn't see the direct application :/ – Quixotic Aug 26 '11 at 17:40
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    @Foo One key thing to keep in mind is that radix notation has polynomial form, so this can be viewed as a special case of well-known results about polynomials. Always be on the lookout for analogies between numbers and functions (here polynomials). – Bill Dubuque Aug 26 '11 at 17:55
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HINT $\ $ Employ the radix $\:9\:$ analog of casting out nines in decimal, namely

$$\rm\quad mod\ 8:\ \ \ 9 \equiv 1\ \ \Rightarrow\ \ d_n\: 9^n +\:\cdots\:+d_1\:9 + d_0\ \equiv\ d_n +\:\cdots\:+d_1+d_0$$

The same idea works generally to cast out $\rm\:b\pm1\:$'s in radix $\rm\:b\:$, e.g. see here for many links.

It may be viewed as a number-theoretic specialization of the Polynomial Remainder Theorem

$\rm\quad f(x)\ mod\ (x-r)\:=\:f(r)\:,\ $ thus $\rm\ \ f(x)\ mod\ (x-1)\:=\:f(1)\: =\: f_n +\:\cdots\:+f_1 + f_0$

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Bill Dubuque
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