2001 AIME II Problems/Problem 10

Problem

How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$?

Solution 1

The prime factorization of $1001 = 7\times 11\times 13$. We have $7\times 11\times 13\times k = 10^j - 10^i = 10^i(10^{j - i} - 1)$. Since $\text{gcd}\,(10^i = 2^i \times 5^i, 7 \times 11 \times 13) = 1$, we require that $1001 = 10^3 + 1 | 10^{j-i} - 1$. From the factorization $10^6 - 1 = (10^3 + 1)(10^{3} - 1)$, we see that $j-i = 6$ works; also, $a-b | a^n - b^n$ implies that $10^{6} - 1 | 10^{6k} - 1$, and so any $\boxed{j-i \equiv 0 \pmod{6}}$ will work.

To show that no other possibilities work, suppose $j-i \equiv a \pmod{6},\ 1 \le a \le 5$, and let $j-i-a = 6k$. Then we can write $10^{j-i} - 1 = 10^{a} (10^{6k} - 1) + (10^{a} - 1)$, and we can easily verify that $10^6 - 1 \nmid 10^a - 1$ for $1 \le a \le 5$.

If $j - i = 6, j\leq 99$, then we can have solutions of $10^6 - 10^0, 10^7 - 10^1, \dots\implies 94$ ways. If $j - i = 12$, we can have the solutions of $10^{12} - 10^{0},\dots\implies 94 - 6 = 88$, and so forth. Therefore, the answer is $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$.

Solution 2

Observation: We see that there is a pattern with $10^k \pmod{1001}$. \[10^0 \equiv 1 \pmod{1001}\] \[10^1 \equiv 10 \pmod{1001}\] \[10^2 \equiv 100 \pmod{1001}\] \[10^3 \equiv -1 \pmod{1001}\] \[10^4 \equiv -10 \pmod{1001}\] \[10^5 \equiv -100 \pmod{1001}\] \[10^6 \equiv 1 \pmod{1001}\] \[10^7 \equiv 10 \pmod{1001}\] \[10^8 \equiv 100 \pmod{1001}\]

So, this pattern repeats every 6.

Also, $10^j-10^i \equiv 0 \pmod{1001}$, so $10^j \equiv 10^i \pmod{1001}$, and thus, \[j \equiv i \pmod{6}\]. Continue with the 2rd paragraph of solution 1, and we get the answer of $\boxed{784}$

-AlexLikeMath

See also

2001 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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