Hi I have been working on this problem and I don't understand my textbook's solution to this problem.

Here is my textbook's solution to the problem: 

We have 1001 = 7\*11\*13 and the condition of the problem says 7\*11\*13\*k = 10^j - 10^i = (10^i)(10^(j-I) - 1). This implies 10^j-i = 1 (mod 1001) (and also (mod 7)). Since phi(7) = 6, by Euler's totient theorem, j-I = 0 (mod 6). If j-I = 6 we have 94 possibilities for j-I: 6-0, 7-1, ..., 99-93. If j-I = 12 we have 94-6 = 88 possibilities for j-I: 12-0, 13-1, ... , 99-87. The last case is j-I = 96 where we have 4 possibilities: 96-0, 97-1, 98-2, and 99-3. In total we have 94+88+82+...+10+4 = 16 \* (98/2) = 784 possibilities.

I don't get how (10^i)(10^(j-I) - 1) implies 10^(j-I) = 1 (mod 1001) and (mod 7), or why my textbook used the totient function on 7, instead of 1001. I also don't quite understand their method for finding the different possibilities for j-I.
['https://math.stackexchange.com/questions/1890332', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/360993/']
You have $7 \cdot 11 \cdot 13 \cdot k$ is a multiple of $1001$. You know $10^I$ is not a multiple of $1001$, so the other factor $10^{j-I}-1$ must be a multiple of $1001$ and hence of all its factors. The rest of the proof is incorrect but results in the correct answer. Euler's totient theorem says that $a^{\phi(n)} \equiv 1 \pmod n$ for $a$ coprime to $n$. As $7$ is prime, $\phi(7)=6$, so if $j-I$ is a multiple of $6, 10^{j-I}\equiv 1 \pmod 7$ What they say is $10^{j-I}\equiv 1 \pmod 7$ proves that $j-I$ must be a multiple of $6$. The book is reading the implication in the wrong direction. We can see the problem by looking at another factor of $1001,$ which is $11$. The book argument would also claim that we must have $j-I$ a multiple of $10$, but in fact working $\bmod 11$ only tells us that $j-I$ is even, which we knew already. We know that $10^6 \equiv 3^6 \equiv 1 \pmod 7$ but nothing in the argument shows that no lower power of $10$ is also $\equiv 1 \pmod 7$. That is what happens for $11$ and $13$, so all cases where $6|(j-I)$ result in $10^j-10^I$ being a multiple of $1001$
Forgive the second answer but you book just gives the wrong reason.

Instead here's the "pretty picture" way:

$10^j - 10^i = 100000..... - 1000.... = 99999999.....90000.....$

where there are $j-i$ 9's and $i$ zeros. This is divisble by $1001$ only if $999...9 = 10^{j-i} -1$ is.

So take some pencil and paper and do the problem $99999.... \div 1001$. If you do long division (try it) you get: $999000999000.....$ and this only divides evenly if and only if $j - i$ is a multiple of $6$.

But that's a very informal proof. 

More formally we can do what your text does. $10^j - 10^i = (10^{j-i} - 1)10^i$. As $\gcd(10^i, 1001) = 1$, $1001|10^j - 10^i \iff 1001|10^{j-i}-1$.

Your text shows that $1001|10^{j-i} - 1 \implies 10^{j-i} \equiv 1 \mod (7|11|13)$ But that's not enough to go the other way. The text is dead wrong about that.

But let $k = j-i$. If $6|k$ so $k = 6n$ then we have $10^{6n} - 1 = (10^3+1)(10^{6k -3} - 10^{6k - 6}+....+10^9 - 10^6 + 10^3 - 1)= (1001)\*\sum\_{l=0}^{2k-1}(-1)^{l+1}10^{3k}$. 

So $10^3 +1|10^{6n} -1$. 

But are these the *only* answers?

Well, yes. If $6 \not \mid k$ then by Fermat's Little theoem if $k \equiv l \mod 6; l \ne 0$ then $10^k \equiv 10^l \not \equiv 1 \mod 7$. So $10^k \not \equiv 1 \mod 1001$ and $1001 \not \mid 10^k - 1$ and $1001 \not \mid 10^j - 10^i$.

Thus the text did get the right answer for the wrong reasons.