Difference between revisions of "2006 AIME II Problems/Problem 14"
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because each digit will appear <math>10^{n - 1}</math> times in each place in the numbers <math>1, 2, \ldots, 10^{n} - 1</math>, and there are <math>n</math> total places. | because each digit will appear <math>10^{n - 1}</math> times in each place in the numbers <math>1, 2, \ldots, 10^{n} - 1</math>, and there are <math>n</math> total places. | ||
− | The denominator of <math>K</math> is <math>D = 2^3\cdot 3^2\cdot 5\cdot 7</math>. For <math>S_n</math> to be an integer, <math>n10^{n-1}</math> must be divisible by <math>D</math>. Since <math>10^{n-1}</math> only contains the factors <math>2</math> and <math>5</math> (but will contain enough of them when <math>n \geq 3</math>), we must choose <math>n</math> to be [[divisible]] by <math>3^2\cdot 7</math>. Since we're looking for the smallest such <math>n</math>, the answer is <math>\boxed{ | + | The denominator of <math>K</math> is <math>D = 2^3\cdot 3^2\cdot 5\cdot 7</math>. For <math>S_n</math> to be an integer, <math>n10^{n-1}</math> must be divisible by <math>D</math>. Since <math>10^{n-1}</math> only contains the factors <math>2</math> and <math>5</math> (but will contain enough of them when <math>n \geq 3</math>), we must choose <math>n</math> to be [[divisible]] by <math>3^2\cdot 7</math>. Since we're looking for the smallest such <math>n</math>, the answer is <math>\boxed{063}</math>. |
== See also == | == See also == |
Latest revision as of 23:32, 28 February 2021
Problem
Let be the sum of the reciprocals of the non-zero digits of the integers from to inclusive. Find the smallest positive integer for which is an integer.
Solution
Let . Examining the terms in , we see that since each digit appears once and 1 appears an extra time. Now consider writing out . Each term of will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so .
In general, we will have that
because each digit will appear times in each place in the numbers , and there are total places.
The denominator of is . For to be an integer, must be divisible by . Since only contains the factors and (but will contain enough of them when ), we must choose to be divisible by . Since we're looking for the smallest such , the answer is .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.