1982 AMC 12 Problem 30

Attempt Problem 30 of the 1982 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1982 AMC 12 solutions, or check the answer key.

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30.

Find the units digit in the decimal expansion of (15+220)19+(15+220)82. (15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}.
00
22
55
99
none of these
Answer: D
Concepts:recursionmodular arithmeticunits digit
Small hint:
Pair α=15+220\alpha=15+\sqrt{220} with its conjugate β=15−220\beta=15-\sqrt{220}
Big hint:
The integers Sn=αn+βnS_n=\alpha^n+\beta^n satisfy a short recurrence, while 0<β<10\lt\beta\lt1
Solution:
Let α=15+220\alpha=15+\sqrt{220} and β=15−220,\beta=15-\sqrt{220}, so 0<β<1.0\lt\beta\lt1. The integers Sn=αn+βnS_n=\alpha^n+\beta^n satisfy Sn=30Sn−1−5Sn−2,S_n=30S_{n-1}-5S_{n-2}, hence Sn≡0(mod10)S_n\equiv0\pmod{10} for every n≥1.n\ge1. Therefore α19+α82=S19+S82−(β19+β82) \begin{aligned} \alpha^{19}+\alpha^{82} &=S_{19}+S_{82}\\ &\quad-(\beta^{19}+\beta^{82}) \end{aligned} is a multiple of 1010 minus a positive number less than 1.1. Its integer part therefore ends in 9.9.
Therefore, the correct answer is D.
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