Big Division

Let $x=\underbrace{2323\dots2323}_{\text{4018 digits}}$ . To find the remainder when $x$ is divided by 999, we reduce the number of digits by 6 each time. We can do this as follows:

First we note the following: $\begin{aligned} \underbrace{2323\dots2323}_{n \text{ digits}} &= \underbrace{2323...2323}_{n-6\text{ digits}} \times10^6+ 232323 \\ &=\underbrace{2323...2323}_{n-6\text{ digits}} \times 999999 + \underbrace{2323...2323}_{n-6\text{ digits}} + 232323 \end{aligned}$ We apply modulo 999 to both sides. Since 999999 is divisible by 999 and 232323 modulo 999 is 555 we get: $\begin{aligned} \underbrace{2323\dots2323}_{n \text{ digits}} \text{ mod } 999 &= \underbrace{2323...2323}_{n-6\text{ digits}} \text{ mod } 999 + 232323 \text{ mod }999 \\ &= \underbrace{2323...2323}_{n-6\text{ digits}} \text{ mod }999 + 555 \end{aligned}$

Now we apply this repeatedly to the original number $x$ . Since we are repeatedly reducing the number of digits by 6, the above process can be applied $4018 \div 6$ times, i.e. 669 times. The remainder of 4018 divided by 6 is 4, and so we will be left with 2323 at the end. Thus we get

$\begin{aligned} x \text{ mod } 999 &= (2323 + 555 \times 669 ) \text { mod } 999 \\ &= (2323 + 555 \times 666 + 555 \times 3 ) \text { mod } 999 \\ &= (2323 + 1665) = \boxed{991} \text { mod } 999 \end{aligned}$

Since $555 \times 666$ is divisible by 999, $x$ modulo 999 is equal to 2323+1665 modulo 999. The remainder of 2323+1665, i.e. 3988 when divided by 999 is 991 which is our final answer.

We all know how to test whether an integer is divisible by 3 or 9. My solution relies on knowing that the method can be generalised on two levels:

• Every non-negative integer is congruent to its decimal digit sum (and hence digital root) modulo 3 or 9.
• For arbitrary base $n$ , the moduli on which this works are the factors of $n - 1$ .

As such, all we need to do is calculate the base-1000 digital root of $x$ .

Now $4018 = 6 \times 669 + 4 \\ \Rightarrow x = 2\ 323\ \underbrace{232\ 323}_{\text{669 times}} \\ \Rightarrow x \equiv 2 + 323 + 669(232 + 323) \text{ (mod 999)} \\ = 371620 \\ \equiv 371 + 620 \text{ (mod 999)} \\ = \boxed{991}.$

Since $10^{3}\equiv{1}\pmod{999},$ we obtain that $10^{6n}\equiv{1}\pmod{999}\:\:\:\:\: (*)$ for any integer $n.$ Then $N=\underbrace {232323...232323}_{4018\text{ digits}}=23*\underbrace{10101...010101}_{4017 \text{ digits}}$ $=23* (10101*\sum_{k=0}^{668}10^{6k}+10^{4014}+10^{4016})$ Then $N=23* (10101*\sum_{k=0}^{668}10^{6k}+10^{6*669}+100*10^{6*669}).$

Now using $(*)$ we obtain that $N\equiv 23*(10101*669+1+100)\equiv 991 \pmod{999}.$ Therefore the answer is $991.$

$\underbrace{232323\ldots232323}_{4018 \text{ digits}} = \displaystyle \sum_{k = 0}^{2008} 23\cdot 10^{2k} \\ 10^0 \equiv 10^3 \equiv 10^{3n} \equiv 1 \pmod {999} \Rightarrow \boxed{23\cdot 10^{3n} \equiv 23 \pmod {999}} (1) \\ \Rightarrow 10\cdot 23\cdot 10^{3n} \equiv 10\cdot 23 \pmod {999} \Rightarrow \boxed{23\cdot 10^{3n +1} \equiv 230 \pmod {999}} (2)\\ \Rightarrow 23\cdot 10^{3n +2} \equiv 2300 \equiv 302 \pmod {999} \Rightarrow \boxed{23\cdot 10^{3n +2} \equiv 302 \pmod {999}} (3) \\ \text{The problem is equivalent to the following congruence: } \\ \displaystyle \sum_{k = 0}^{2008} 23\cdot 10^{2k} \equiv {?} \pmod{999} \\ \text{Every single term of the above series is either equivalent to (1) or (2) or (3)} \\ \displaystyle \sum_{k = 0}^{2008} 23\cdot 10^{2k} \equiv \underbrace{23 + 302 + 230 +\ldots + 23 + 302 + 230}_{2007 \text{ terms}} + 23 + 302 \pmod{999} \\ \equiv \frac{2007}{3}\cdot (23 + 302 + 230) + 23 + 302 \equiv 371620 \equiv - 8 \equiv \boxed{991 \pmod{999}}$

We can write the given number in blocks of three, as one has a habit of doing (in business, for example): x = 2'323'...'232'323. There are 670 blocks of 323s, 669 blocks of 232s, and one "left-over block" of just a 2. Since $10^{3n}\equiv{1}\pmod{999}$ , the given number is congruent to the sum of the blocks, $670\times{323}+669\times{232}+2$ $\equiv{\boxed{991}}\pmod{999}$

step one: mod 37 of the series follows 0,23,29 after six digits(232323) with a pair addition of one '23'..giving mod 37 of the number=29

step two: divisibility test for 27,sum of groups of three numbers from right to be divisible by 27, .. 669(4017/3) times groups of 323 and 232 and last block of 323 to adjust for the sum to be divided by 27,hence getting (2+323x669+232x669+304) is divisible by 27 hence mod 27 of the actual number to be 19 (323-304)

Find some p and q for which 37p+29=27q+19 would hold got 37x26+29=27x36+19=991 and thats the remainder.

I, unfortunately, have no training in number theory, so I went about it the hard way. I quickly found a pattern, however.

So, I looked at remainders for the nth digit: N = 3 is 232, N = 4 is 325, N= 5 is 255, ...

The two numbers that helped point me in the right direction were N = 12, and N =15.

N = 12 yields a remainder of 111. N = 15 is 343.

I made a leap from here-- N = 15 is 343, N = 3 is 232. 15-3 = 12. 343-232 = 111

I then hypothesized that: N = 24 would be 222, N = 36 would be 333, etc... til N = 108 would be 999.

It was simple computation that showed me I was right (up to 48). I then calculated how many times 108 goes into 4018 (37 R 22). Which tells me there would be 22 digits of 232323, etc....

Then, as I had already calculated the remainders up to 48, I found the remainder for the 22nd digit which was 991. I took a deep breath and was right!

I have already looked at the other solutions and they make some sense to me, but I don't have a strong enough background in number theory to completely understand what they mean.

My way was likely longer and more tedious than others, but it helped me develop a better intuition for division and patterns.

I pretty much followed the obvious "base 1000" approach that everyone else did, noting that 1000 = 1 (mod 999) so that the remainder mod 999 is the same as the sum of all groups of 3 digits, counting from the right: 2 + 323 + (232 + 323)*669.

I'm curious, though. Did Calvin pick the numbers such that the "hard" part was:

(232 + 323)*669 = 666 (mod 999)

...with "the number of the beast" both right-side up and upside down?

Let $x$ be the given dividend and $y = 23\cdot 10^{4018}+x$ . So $y$ is a concatenation of 670 strings of "232323.". Since, under modulo 999, $10^{3k} = 1$ , for all integer $k >= 0$ , so under modulo 999, $y = 670\cdot 232323 = 670\cdot (232+323) = 670\cdot 555 = (666+4)\cdot 555 = 10\cdot 111\cdot333+4\cdot 555 = 10\cdot 37\cdot 999+2220 = 0+2+220 = 222,$ and so $x = 222-23\cdot 10^{4018} = 222-230 = -8 = 991.$ Hence the desired remainder is 991.

We may write the given number as $23\sum_{n=0}^{2008} 100^n$

Trying any normal methods like geometric series or applying Euler's theorem does not work (though an application of Euler's theorem simplifies my argument but as it is not necessary I omit it). This immediately forces us to ask what is the deal with 999. Well, obviously $1000 \equiv 1 \mod 999$ which means that in numbers of the form $1000...$ we may delete strings of 000s. In other words, $\sum_{n=0}^{2008} 100^n \equiv 1 + 100 + 10 + 1 + 100 + 10 + ... \mod 999$ .

We must now just formalize this to obtain an answer. As the powers of 100 have a cycle of length 3 compute $2008 \equiv 1 \mod 3$ and $\lfloor \frac{2008}{3} \rfloor = 669$ . This means that $\sum_{n=0}^{2008} 100^n \equiv 111 \times 699 + 1 + 100 \equiv 434 \mod 999$ . And multiplying, $23\sum_{n=0}^{2008} 100^n \equiv 23 \times 434 \equiv 991 \mod 999$ .

The easiest way to solve this problem $a_{n}a_{n-1} ... a_{1}a_{0} \equiv{ \sum_{k=0}^{n}a_{k}*10^{(k\mod{3)}}} \pmod{999}$ In our case this gives $\underbrace{232323...232323}_{4018 \text{ digits}} \equiv{371620} \equiv{991} \pmod{999}$

2323232323....2331:3=77441077441 ...077441777 and this number is divided by 9 because (7+7+4+4+1) 669+21= 3 (223 23+7)=3(50676)=9 16892 So x+8 is divided by 27 23232323....2331:37= 6279|6279|…6279|63 (27,37)=1 So x+8 is divided by 27*37=999, so x=999-8=991 (mod 999)