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1978 AHSME Problems/Problem 14: Difference between revisions

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== Problem 14 ==


If an integer <math>n > 8</math> is a solution of the equation <math>x^2 - ax+b=0</math> and the representation of <math>a</math> in the base-<math>n</math> number system is <math>18</math>,
then the base-n representation of <math>b</math> is
<math>\textbf{(A)}\ 18 \qquad
\textbf{(B)}\ 20 \qquad
\textbf{(C)}\ 80 \qquad
\textbf{(D)}\ 81 \qquad
\textbf{(E)}\ 280    </math>
== Solution ==


Assuming the solutions to the equation are n and m, by Vieta's formulas, <math>n_n + m_n = 18_n</math>.   
Assuming the solutions to the equation are n and m, by Vieta's formulas, <math>n_n + m_n = 18_n</math>.   
Line 11: Line 21:


The answer is (C) <math>80</math>
The answer is (C) <math>80</math>
==See Also==
{{AHSME box|year=1978|num-b=13|num-a=15}}
{{MAA Notice}}

Latest revision as of 20:08, 13 February 2021

Problem 14

If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base-$n$ number system is $18$, then the base-n representation of $b$ is

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$

Solution

Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$.

$n_n = 10_n$, so $10_n + m_n = 18_n$.

\[m_n = 8_n\].

Also by Vieta's formulas, $n_n \cdot m_n = b_n$. \[10_n \cdot 8_n = \boxed{80_n}\].

The answer is (C) $80$


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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
All AHSME Problems and Solutions

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