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2001 AIME II Problems/Problem 10

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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}.

See also

2001 AIME II (ProblemsResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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