AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
Personal tools

2001 AIME II Problems/Problem 10

From AoPSWiki

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
Stay informed about new Art of Problem Solving developments.
Click here to join our mailing lists.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us