2006 AIME II Problems/Problem 14
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 |
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