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1987 AJHSME Problems/Problem 21

Revision as of 22:42, 13 March 2009 by 5849206328x (talk | contribs) (New page: ==Problem== Suppose <math>n^{*}</math> means <math>\frac{1}{n}</math>, the reciprocal of <math>n</math>. For example, <math>5^{*}=\frac{1}{5}</math>. How many of the following statement...)
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Problem

Suppose $n^{*}$ means $\frac{1}{n}$, the reciprocal of $n$. For example, $5^{*}=\frac{1}{5}$. How many of the following statements are true? 

i) $3^*+6^*=9^*$
ii) $6^*-4^*=2^*$
iii) $2^*\cdot 6^*=12^*$
iv) $10^*\div 2^* =5^*$

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

Solution

We can just test all of these statements: \begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac{1}{5} = 5^* \end{align*}

The last two statements are true and the first two aren't, so $\boxed{\text{C}}$

See Also

1987 AJHSME Problems

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