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

Answer

Answer: 428

Concepts:binomial theorem modular exponentiation repeating decimal

Difficulty rating: 2640

Solution:

Apply the expansion with $x = 10^{2002},$ $y = 1,$ and $r = \frac{10}{7}:$ $\begin{aligned} &\left(10^{2002} + 1\right)^{10/7} \\ &= 10^{2860} + \frac{10}{7} \cdot 10^{858} \\ &{}+ \frac{\frac{10}{7} \cdot \frac{3}{7}}{2} \cdot 10^{-1144} \\ &{}+ \cdots. \end{aligned}$ 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 $428.$

2002 AIME I Problem 7

Answer: 428

Difficulty rating: 2640

Solution:

Problem 7 in Other Years