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Difference between revisions of "2001 AIME II Problems/Problem 10"
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== Solution ==
 
== Solution ==
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The [[prime factorization]] of <math>1001 = 7\times 11\times 13</math>. We have <math>7\times 11\times 13\times k = 10^j - 10^i = 10^i(10^{j - i} - 1)</math>. Since <math>\text{gcd}\,(10^i = 2^i \times 5^i, 3 \times 7 \times 13) = 1</math>, we require that <math>1001 = 10^3 + 1 | 10^{j-i} - 1</math>. From the factorization <math>10^6 - 1 = (10^3 + 1)(10^{3} - 1)</math>, we see that <math>j-i = 6</math> works; also, <math>a-b | a^n - b^n</math> implies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work.  
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The [[prime factorization]] of <math>1001 = 7\times 11\times 13</math>. We have <math>7\times 11\times 13\times k = 10^j - 10^i = 10^i(10^{j - i} - 1)</math>. Since <math>\text{gcd}\,(10^i = 2^i \times 5^i, 7 \times 11 \times 13) = 1</math>, we require that <math>1001 = 10^3 + 1 | 10^{j-i} - 1</math>. From the factorization <math>10^6 - 1 = (10^3 + 1)(10^{3} - 1)</math>, we see that <math>j-i = 6</math> works; also, <math>a-b | a^n - b^n</math> implies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work.  
  
 
<!-- I cannot make sense of this, so I commented it: - azjps - we require that <math>10^{j - i} - 1\equiv 0 \mod 1001</math>. By [[Euler's totient theorem]], <math>j - i\equiv 0\mod 6</math>. -->
 
<!-- I cannot make sense of this, so I commented it: - azjps - we require that <math>10^{j - i} - 1\equiv 0 \mod 1001</math>. By [[Euler's totient theorem]], <math>j - i\equiv 0\mod 6</math>. -->

Revision as of 12:13, 3 October 2018

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 $\text{gcd}\,(10^i = 2^i \times 5^i, 7 \times 11 \times 13) = 1$, 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 $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
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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