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Difference between revisions of "1978 AHSME Problems/Problem 14"

(Created page with " Assuming the solutions to the equation are n and m, by Vieta's formulas, <math>n_n + m_n = 18_n</math>. <math>n_n = 10_n</math>, so <math>10_n + m_n = 18_n</math>. <cm...")
 
 
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== Problem 14 ==
  
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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>,
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then the base-n representation of <math>b</math> is
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<math>\textbf{(A)}\ 18 \qquad
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\textbf{(B)}\ 20 \qquad
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\textbf{(C)}\ 80 \qquad
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\textbf{(D)}\ 81 \qquad
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\textbf{(E)}\ 280    </math>
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== 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>.   
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The answer is (C) <math>80</math>
 
The answer is (C) <math>80</math>
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==See Also==
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{{AHSME box|year=1978|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 21:08, 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 (Problems • Answer Key • Resources)
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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