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

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 $\text{gcd}\,(10^i = 2^i \times 5^i, 3 \times 7 \times 13) = 1$ , 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 $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2001 AIME II Problems/Problem 10

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