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Finding remainder when $\displaystyle \underbrace{11111111\cdots \cdots 1111}_{\text{1862 times} }$ is divided by $2006$

What i tried ::

We can write it as $\displaystyle 111111111111\cdots \text{1862 times }=1+10+10^2+10^3+10^4+\cdots +10^{1861}=\frac{10^{1862}-1}{10-1}=\frac{10^{1862}-1}{9}$

Also $2006=2\times 17\times 59$

How do i solve it, please Help me , Thanks

Jyrki Lahtonen
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jacky
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  • I think you will need some number theory ingredients. My feeling is to look at the "Chinese Remainder Theorem" and perhaps also "Euler's Theorem". The first you can google as is, and the second one is the theorem mentioning the so-called "totient" function, if you need more to go on in the google search. Even better would be if you had a good book on introductory Number Theory. – coffeemath Jun 25 '22 at 06:39
  • Next you'll want to compute $10^{1862}-1$ modulo $2$ and $9$ and $17$ and $59$, so that you can combine them to compute $10^{1862}-1$ modulo $2\times9\times17\times59$. – Greg Martin Jun 25 '22 at 06:41
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    Hint: $\ 1862\equiv 6,\pmod{!\phi(2006)}\ $ so the result follows immediately by Euler's theorem and mod order reduction in the linked dupe, and the congruence quotient rule. – Bill Dubuque Jun 25 '22 at 07:43
  • Repunits (long strings of $1$s) have some neat divisibility properties. Take as an example $111111$. It's pretty clear this is divisible by both $11$ and $111$: $$111111=11 \cdot 10101=111 \cdot 1001$$ Note that $6=2 \cdot 3$, which tells us which smaller repunits evenly divide a repunits with $6$ $1$s. Cool, huh? Now $1862=2 \cdot 7 \cdot 7 \cdot 19$. So we could write it as $(1111111111111111111)(1+10^{19}+10^{38}+\cdots +10^{(19)(97)})$. Now this won't lead directly to an answer as Euler's theorem (mentioned above) will, but it may do you good, or at least be interesting, just to explore. – Eric Snyder Jun 25 '22 at 08:01
  • @GregMartin $10^n -1 \equiv 0 \pmod 9$ for all $n$, so that pathway sadly leads to ruin. – Eric Snyder Jun 25 '22 at 08:11
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    @Eric Likely Greg intends to scale the modulus by $,9,$ so we we get a congruence to the sought modulus when we later divide by $,9,,$ i.e. $10^k-1 = 9x \equiv a \pmod{!9n} ,\Rightarrow, x \equiv a/9 \pmod{! n}., $ But such modulus scaling is not needed here since $,9,$ is coprime to $,n,$ so invertible $!\bmod n,,$ so we can use $\rm\color{#c00}{Q}$ = congruence quotient rule (i.e. mod fraction reduction) and mod $\rm\color{#0a0}{order}$ reduction as below – Bill Dubuque Jun 25 '22 at 11:03
  • $$\bmod n!=!2006!:,\ \overbrace{\dfrac{10^{\color{#0a0}{1862}}-1}9,\overset{\rm\color{#c00}{Q}}\equiv, \dfrac{10^{\color{#0a0}6}-1}9}^{\large\color{#0a0}{1862,\ \equiv,\ 6}\pmod{!\phi(n)}!!},\equiv, 111111,\equiv, 781\qquad$$ – Bill Dubuque Jun 25 '22 at 11:03
  • @BillDubuque I see what you're saying. I was thinking the intent was to rearrange to $9N+1 \equiv 10^{1862} \equiv 1 ; 000 ; 000 \pmod{2006}$, giving $9N \equiv 1011 \implies N \equiv 781 \pmod{2006}$. – Eric Snyder Jun 30 '22 at 02:41

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