Difference between revisions of "2001 AIME II Problems/Problem 10"

Revision as of 02:55, 12 November 2022

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 1

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 $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$ .

Solution 2

Observation: We see that there is a pattern with $10^k \pmod{1001}$ . $10^0 \equiv 1 \pmod{1001}$ $10^1 \equiv 10 \pmod{1001}$ $10^2 \equiv 100 \pmod{1001}$ $10^3 \equiv -1 \pmod{1001}$ $10^4 \equiv -10 \pmod{1001}$ $10^5 \equiv -100 \pmod{1001}$ $10^6 \equiv 1 \pmod{1001}$ $10^7 \equiv 10 \pmod{1001}$ $10^8 \equiv 100 \pmod{1001}$

So, this pattern repeats every 6.

Also, $10^j-10^i \equiv 0 \pmod{1001}$ , so $10^j \equiv 10^i \pmod{1001}$ , and thus, $j \equiv i \pmod{6}$ . Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$

-AlexLikeMath

Solution 3

Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$ ; therefore, the order of 10 modulo $7,11$ , and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$ .

~Dhillonr25

@@ Line 28: / Line 28: @@
 -AlexLikeMath
+==Solution 3==
+Note that <math>1001=7\cdot 11\cdot 13,</math> and note that <math>10^3 \equiv \pmod{p}</math> for prime <math>p | 1001</math>; therefore, the order of 10 modulo <math>7,11</math>, and <math>13</math> must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that <math>i-j=6k</math> for some natural number k. From here, we note that for <math>j=0,1,2,3,</math> we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is <math>4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}</math>.
+~Dhillonr25
 == 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
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