# 2008 iTest Problems/Problem 61

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## Problem

Find the units digits in the decimal expansion of $\left(2008+\sqrt{4032000}\right)^{2000}+\left(2008+\sqrt{4032000}\right)^{2001}+\left(2008+\sqrt{4032000}\right)^{2002}+\cdots$ $+\left(2008+\sqrt{4032000}\right)^{2007}+\left(2008+\sqrt{4032000}\right)^{2008}$

## Solution

Let $n$ be a positive integer. Note that $2008^2 = 4032064,$ which is close to $4032000.$ That means $(2008-\sqrt{4032000})^{n}$ is close to zero. With this in mind, we find that $(2008+\sqrt{4032000})^{n} + (2008-\sqrt{4032000})^{n}$ is an integer, and the expansion is equal to $\sum_{i=2000}^{2008} \left( (2008+\sqrt{4032000})^{i} + (2008-\sqrt{4032000})^{i} \right) - (2008-\sqrt{4032000})^{i}.$

To find that the units digit of $(2008+\sqrt{4032000})^{n} + (2008-\sqrt{4032000})^{n}$, note that in the expansion of $(2008+\sqrt{4032000})^{n}$ and $(2008-\sqrt{4032000})^{n},$ most of the terms end in a $0.$ That means the units digit equals the units digit of $8^n + 8^n = 2 \cdot 8^n.$

Now we need to find out how small $2008-\sqrt{4032000}$ is. With a calculator, we find that $2008-\sqrt{4032000}=0.016 \text{.}$ But even without a calculator, we can use inequalities to show that $2008-\sqrt{4032000} < \tfrac{1}{10}.$ This means that $\sum_{i=2000}^{2008} (2008-\sqrt{4032000})^i$ would be close to $0.$

Thus, the units digit of the decimal expansion is $2+6+8+4+2+6+8+4+2-1 \rightarrow \boxed{1}.$