Difference between revisions of "2002 AIME I Problems/Problem 7"
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== Problem == | == Problem == | ||
| + | The [[Binomial Expansion]] is valid for exponents that are not integers. That is, for all real numbers <math>x,y</math> and <math>r</math> with <math>|x|>|y|</math>, | ||
| + | |||
| + | <cmath>(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots</cmath> | ||
| + | |||
| + | What are the first three digits to the right of the decimal point in the decimal representation of <math>(10^{2002}+1)^{\frac{10}{7}}</math>? | ||
== Solution == | == Solution == | ||
| − | + | <math>1^n</math> will always be 1, so we can ignore those terms, and using the definition (<math>2002 / 7 = 286</math>): | |
| − | = | + | <cmath>(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots</cmath> |
| − | + | ||
| + | Since the exponent of the <math>10</math> goes down extremely fast, it suffices to consider the first few terms. Also, the <math>10^{2860}</math> term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in | ||
| + | |||
| + | <cmath>\dfrac{10}{7}10^{858}</cmath>. | ||
| + | |||
| + | (The remainder after this term is positive by the [[Remainder Estimation Theorem]]). Since the repeating decimal of <math>\dfrac{10}{7}</math> repeats every 6 digits, we can cut out a lot of 6's from <math>858</math> to reduce the problem to finding the first three digits after the decimal of | ||
| + | |||
| + | <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>\boxed{428}</math>. | ||
| − | + | ---- | |
| + | An equivalent statement is to note that we are looking for <math>1000 \left\{\frac{10^{859}}{7}\right\}</math>, where <math>\{x\} = x - \lfloor x \rfloor</math> is the fractional part of a number. By [[Fermat's Little Theorem]], <math>10^6 \equiv 1 \pmod{7}</math>, so <math>10^{859} \equiv 3^{6 \times 143 + 1} \equiv 3 \pmod{7}</math>; in other words, <math>10^{859}</math> leaves a residue of <math>3</math> after division by <math>7</math>. Then the desired answer is the first three decimal places after <math>\frac 37</math>, which are <math>\boxed{428}</math>. | ||
| − | + | == See also == | |
| + | {{AIME box|year=2002|n=I|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:54, 4 July 2013
Problem
The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers 
and 
with 
,
![\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]](http://latex.artofproblemsolving.com/4/3/c/43c3768ec432dd7b10015400b663d91b7ef8e6aa.png)
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 (
):
![\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]](http://latex.artofproblemsolving.com/f/9/3/f93c3689c3938500497b7e607dd4d966f26d22dc.png)
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
![\[\dfrac{10}{7}10^{858}\]](http://latex.artofproblemsolving.com/d/5/3/d5307a89b95cd1016bba711891618a1b63d7a300.png)
.
(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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
