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Difference between revisions of "1982 AHSME Problems/Problem 30"

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&\equiv 0\pmod{10}.
 
&\equiv 0\pmod{10}.
 
\end{align*}</cmath>
 
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It is clear that <math>0<B^{82}<B^{19}<B<0.5,</math> from which <math>0<B^{19}+B^{82}<0.5+0.5=1.</math> Since the units digit of the decimal expansion of <math>\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)</math> is <math>0,</math> the units digit of the decimal expansion of <math>A^{19}+A^{82}</math> is <math>\boxed{\textbf{(D)}\ 9}.</math>
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It is clear that <math>0<B^{82}<B^{19}<B<0.5,</math> from which <math>0<B^{19}+B^{82}<0.5+0.5=1.</math> We conclude that the units digit of the decimal expansion of <math>B^{19}+B^{82}</math> is <math>0.</math> Since the units digit of the decimal expansion of <math>\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)</math> is <math>0,</math> the units digit of the decimal expansion of <math>A^{19}+A^{82}</math> is <math>\boxed{\textbf{(D)}\ 9}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 00:15, 12 September 2021

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

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 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 $0<B^{19}+B^{82}<0.5+0.5=1.$ 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

See Also

1982 AHSME (Problems • Answer Key • Resources)
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