2002 AIME I Problems/Problem 7

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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 $(10^{2002}+1)^{\frac{10}{7}}$?


$1^n$ 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


(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


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 (ProblemsAnswer KeyResources)
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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