1987 AJHSME Problems/Problem 21

Revision as of 22:53, 4 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS