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1982 AHSME Problems/Problem 30

Problem

Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{none of these}$

Solution 1 (Binomial Expansion)

Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.

We have \begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*} Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\] We add the two equations and take the sum modulo $10:$ \begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*} It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{\textbf{(D)}\ 9}.$

~MRENTHUSIASM

Solution 2 (Characteristic Polynomials)

Let $a_n = \left(15 + \sqrt{220}\right)^n + \left(15 - \sqrt{220}\right)^n$. Since $\left(15 + \sqrt{220}\right) + \left(15 - \sqrt{220}\right) = 30$ and $\left(15 + \sqrt{220}\right)\left(15 - \sqrt{220}\right) = 225 - 220 = 5$, by Vieta's rules, $15 + \sqrt{220}$ and $15 - \sqrt{220}$ are roots of the polynomial $x^2 - 30x + 5$. This must be the characteristic polynomial of $a_n$, and therefore $a_{n} = 30a_{n-1} - 5a_{n-2}$ for all $n \geq 2$.

So $a_0 = 2$, $a_1 = 30$, and $a_2 = 30 \cdot 30 - 5 \cdot 2 = 890$. Since $a_1$ and $a_2$ are divisible by $10$, $a_3$ must also be divisible by $10$. By similar logic, $a_n$ is divisible by $10$ for every integer $n \geq 1$. So $a_{19} + a_{82}$ must be divisible by $10$.

Now $15 - \sqrt{220} < 15 - \sqrt{210.25} = 15 - 14.5 = 0.5$, so $\left(15 - \sqrt{220}\right)^n < (0.5)^n$. Let $x = \left(15 - \sqrt{220}\right)^{19} + \left(15 - \sqrt{220}\right)^{82}$. Then $x < (0.5)^{19} + (0.5)^{82} < 0.5 + 0.5 = 1$.

Therefore, $\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82} = a_{19} + a_{82} - x > a_{19} + a_{82} - 1$, so $\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}$ must be between $a_{19}+a_{82}-1$ and $a_{19}+a_{82}$. Since $a_{19} + a_{82}$ is divisible by $10$, $a_{19}+a_{82}-1$ has a units digit of $9$ and $a_{19}+a_{82}$ has a units digit of $0$. Thus the decimal expansion of $\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}$ has a units digit of $\boxed{(\mathbf{D})\ 9}$.

-j314andrews

See Also

1982 AHSME (Problems • Answer Key • Resources)
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