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

Factoring $1001 = 7\times 11\times 13$ . We have $7\times 11\times 13\times k = 10^j - 10^i\implies \dfrac{7(11)(13)(k)}{10^i} = 10^{j - i} - 1$ . This means that $10^{j - i} - 1\equiv 0 \mod 1001$ . By Euler's totient theorem, $j - i\equiv 0\mod 6$ . If $j - i = 6, j\leq 99$ , then we can have solutions of $10^6 - 10^0, 10^7 - 10^1, \dots\implies 94$ . If $j - i = 12$ , we can have the solutions of $10^{12} - 10^{0},\dots\implies 94 - 6 = 88$ . Likewise, we have $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$ .

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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