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Difference between revisions of "2002 AIME I Problems/Problem 7"

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== Problem ==
 
== Problem ==
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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>,
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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}+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y\cdots</cmath>
+
<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>?
 
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 ==
−
{{solution}}
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<math>1^n</math> will always be 1, so we can ignore those terms, and using the definition (<math>2002 / 7 = 286</math>):
  
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== See also ==
+
<cmath>(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots</cmath>
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* [[2002 AIME I Problems/Problem 6| Previous problem]]
+
 
 +
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>.
  
−
* [[2002 AIME I Problems/Problem 8| Next problem]]
+
----
 +
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>.
  
−
* [[2002 AIME I Problems]]
+
== See also ==
 +
{{AIME box|year=2002|n=I|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 18:54, 4 July 2013

Problem

The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$,

\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]

What are the first three digits to the right of the decimal point in the decimal representation of $(10^{2002}+1)^{\frac{10}{7}}$?

Solution

$1^n$ will always be 1, so we can ignore those terms, and using the definition ($2002 / 7 = 286$):

\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]

Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in

\[\dfrac{10}{7}10^{858}\].

(The remainder after this term is positive by the Remainder Estimation Theorem). Since the repeating decimal of $\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of

$\dfrac{10}{7}$.

That is the same as $1+\dfrac{3}{7}$, and the first three digits after $\dfrac{3}{7}$ are $\boxed{428}$.

An equivalent statement is to note that we are looking for $1000 \left\{\frac{10^{859}}{7}\right\}$, where $\{x\} = x - \lfloor x \rfloor$ is the fractional part of a number. By Fermat's Little Theorem, $10^6 \equiv 1 \pmod{7}$, so $10^{859} \equiv 3^{6 \times 143 + 1} \equiv 3 \pmod{7}$; in other words, $10^{859}$ leaves a residue of $3$ after division by $7$. Then the desired answer is the first three decimal places after $\frac 37$, which are $\boxed{428}$.

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

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