AIME 2002 I · 第 7 题

Solution 1

$1^n$ 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 $\boxed{428}$.

Solution 2

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 $\boxed{428}$.