Difference between revisions of "2006 AIME II Problems/Problem 14"
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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
![$S_n = (n10^{n-1})K + 1$](http://latex.artofproblemsolving.com/d/c/0/dc0713c742f497c8b62f5d736e28233206a1dc9b.png)
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.