# 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 $2+6+8+4+2+6+8+4+2-1 \rightarrow \boxed{1}.$

## See Also

 2008 iTest (Problems) Preceded by:Problem 60 Followed by:Problem 62 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
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