Difference between revisions of "2002 AIME I Problems/Problem 7"
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<math>\dfrac{10}{7}</math>. | <math>\dfrac{10}{7}</math>. | ||
− | That is the same as <math>1+\dfrac{3}{7}</math>, and the first three digits after <math>\dfrac{3}{7}</math> are <math>428</math>. | + | That is the same as <math>1+\dfrac{3}{7}</math>, and the first three digits after <math>\dfrac{3}{7}</math> are <math>\boxed{428}</math>. |
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Revision as of 15:08, 22 October 2011
Problem
The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers and
with
,
What are the first three digits to the right of the decimal point in the decimal representation of ?
Solution
will always be 1, so we can ignore those terms, and using the definition (
):
Since the exponent of the goes down extremely fast, it suffices to consider the first few terms. Also, the
term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in
.
(The remainder after this term is positive by the Remainder Estimation Theorem). Since the repeating decimal of repeats every 6 digits, we can cut out a lot of 6's from
to reduce the problem to finding the first three digits after the decimal of
.
That is the same as , and the first three digits after
are
.
An equivalent statement is to note that we are looking for , where
is the fractional part of a number. By Fermat's Little Theorem,
, so
; in other words,
leaves a residue of
after division by
. Then the desired answer is the first three decimal places after
, which are
.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |