2002 AIME I Problem 7

Question

The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers

x,

y,

and

r

with

|x| \gt |y|,

\begin{aligned} &(x + y)^r \\ &= x^r + r x^{r-1} y \\ &{}+ \frac{r(r - 1)}{2!}\,x^{r-2} y^2 \\ &{}+ \frac{r(r - 1)(r - 2)}{3!}\,x^{r-3} y^3 \\ &{}+ \cdots \end{aligned}

What are the first three digits to the right of the decimal point in the decimal representation of

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

Answer

Expand with

x = 10^{2002},

y = 1,

r = \frac{10}{7}:

the first term is an integer and the third term onward is negligibly small

The requested digits form the fractional part of

\frac{10^{859}}{7};

use the period-

6

repeating decimal of

\frac{1}{7}

Apply the expansion with

x = 10^{2002},

y = 1,

and

r = \frac{10}{7}:

\begin{aligned} &\left(10^{2002} + 1\right)^{\frac{10}{7}} \\ &= 10^{2860} + \frac{10}{7} \cdot 10^{858} \\ &{}+ \frac{\frac{10}{7} \cdot \frac{3}{7}}{2} \cdot 10^{-1144} \\ &{}+ \cdots. \end{aligned}

The first term is an integer, and the third and later terms are far smaller than

10^{-1000},

too small to affect the leading decimal digits. So those digits come from the fractional part of

\frac{10}{7} \cdot 10^{858} = \frac{10^{859}}{7}.

That fractional part is

\frac{10^{859} \bmod 7}{7}.

Since

10^6 \equiv 1 \pmod{7}

and

859 \equiv 1 \pmod{6},

we get

10^{859} \equiv 10 \equiv 3 \pmod{7},

so the fractional part is

\frac{3}{7} = 0.428571\ldots

The first three digits to the right of the decimal point are

428.

Problem 7 in other years