1977 AHSME Problems/Problem 11

Problem 11

For each real number $x$ , let $\textbf{[}x\textbf{]}$ be the largest integer not exceeding $x$ (i.e., the integer $n$ such that $n\le x<n+1$ ). Which of the following statements is (are) true?

$\textbf{I. [}x+1\textbf{]}=\textbf{[}x\textbf{]}+1\text{ for all }x \\ \textbf{II. [}x+y\textbf{]}=\textbf{[}x\textbf{]}+\textbf{[}y\textbf{]}\text{ for all }x\text{ and }y \\ \textbf{III. [}xy\textbf{]}=\textbf{[}x\textbf{]}\textbf{[}y\textbf{]}\text{ for all }x\text{ and }y$

$\textbf{(A) }\text{none}\qquad \textbf{(B) }\textbf{I }\text{only}\qquad \textbf{(C) }\textbf{I}\text{ and }\textbf{II}\text{ only}\qquad \textbf{(D) }\textbf{III }\text{only}\qquad \textbf{(E) }\text{all}$

Solution

Solution by e_power_pi_times_i

Notice that $\textbf{[}x\textbf{]}$ is just $\left \lfloor x \right \rfloor$ . We see that $\textbf{I}$ is true, as adding by one does not change the fraction part of the number. Similarly, $\textbf{II}$ is false, because $\textbf{[}x+y\textbf{]}$ does not always equal $\textbf{[}x\textbf{]}+\textbf{[}y\textbf{]}$ (If both numbers were $1.5$ , $\textbf{[}1.5+1.5\textbf{]} = 3$ , $\textbf{[}1.5\textbf{]}+\textbf{[}1.5\textbf{]} = 2$ ). Because $\textbf{I}$ is true and $\textbf{II}$ is false, the answer is $\boxed{\textbf{(B) }\textbf{I }\text{only}}$ .

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1977 AHSME Problems/Problem 11

Problem 11

Solution