It`s srtike me as a little odd ! http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150536

WakeUp wrote: Prove that there are at least $666$ positive composite numbers with $2006$ digits, having a digit equal to $7$ and all the rest equal to $1$. The $334$ numbers $\frac{10^{2006}-1}9+6\cdot10^{6k+4}$ for $k=0\to 333$ all are divisible by $7$ The $334$ numbers $\frac{10^{2006}-1}9+6\cdot10^{6k+2}$ for $k=0\to 333$ all are divisible by $13$

What is the motivation to find that?

harry1234 wrote: What is the motivation to find that? Just direct application of the problem itself. We are looking for at least $666$ pairs $(k,p)$ where $k\in\{0,1,2,...,2005\}$ and $p$ prime such that $\frac{10^{2006}-1}9+6.10^k\equiv 0\pmod p$ Avoiding $p=3$ (in order to be allowed to multiply by $9$), this is equivalent to $10^{2006}+54.10^k\equiv 1\pmod p$ $p=5$ is obviously not an useful candidate. So the first trial is $p=7$ It is easy to see that $\pmod 7$, $10^k$ is $1,3,2,6,4,5,...,1,3,2,6,4,5,...$ So $10^{2006}\equiv 2\pmod 7$ and equation is $10^k\equiv 4\pmod 7$ and so $k=6n+4$ And $6n+4\in\{0,1,...,2005\}$ means $n\in\{0,333\}$ Second trial is $p=11$ but equation becomes $10^k\equiv 0\pmod{11}$, useless Third trial is $p=13$ It is easy to see that $\pmod {13}$, $10^k$ is $1,10,9,12,3,4,...,1,10,9,12,3,4,,...$ So $10^{2006}\equiv 9\pmod {13}$ and equation is $10^k\equiv 9\pmod {13}$ and so $k=6n+2$ And $6n+2\in\{0,1,...,2005\}$ means $n\in\{0,333\}$

It is easy to see that the numbers of this kind will not be divisible by 2, 3, or 5. So we test 7. The $334$ numbers $\frac{10^{2006}-1}9+6*10^{6n+4}$ for $n=[0,333]$ all are divisible by $7$, because $6*10^{6n+4}\equiv 6*10^4*(10^6)^n\equiv -10000 \equiv 3\mod{7}$. Then it suffices to say that $\frac{10^{2006}-1}{9}$ is 4 mod 7. Indeed, we have $10^{2006}\equiv 10^{2006 \mod{6}}\equiv 100\equiv 37=1+9*(7-3)\mod{7}$ concludes that it is $0\mod7$. Similarly for 13 (11 clearly does not need trying), The $334$ numbers $\frac{10^{2006}-1}9+6*10^{6n+2}$ for $n=[0,333]$ all are divisible by $13$. This makes 668$\geq$666 numbers.

Note that none of these numbers will be divisible by $2,3,5$ so we test $7$, the $334$ numbers in the form of $\frac{10^{2006}-1}{9}+6\cdot10^{6n+4}$ when $n$ is between $0$ and $333$ are divisible by 7, and the $334$ numbers in the form of $\frac{10^{2006}-1}{9}+6\cdot10^{6n+2}$ when $n$ is between $0$ and $333$ are divisible by $13$, hence proved.