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If \mathrm{x, y \in(0,30)} such that
\mathrm{\left[\frac{x}{3}\right]+\left[\frac{3 x}{2}\right]+\left[\frac{y}{2}\right]+\left[\frac{3 y}{4}\right]=\frac{11}{6} x+\frac{5}{4} y}

(where [x] denotes greatest integer \mathrm{\leq x} ), then number of ordered pairs \mathrm{(x, y)}is

Option: 1

0


Option: 2

2


Option: 3

4


Option: 4

none of these


Answers (1)

best_answer

Let \mathrm{\{x\}=x-[x]} denote the fractional part of \mathrm{x}.

Note that \mathrm{0 \leq\{x\}<1}. We can write the given equation as

\mathrm{\frac{x}{3}-\left\{\frac{x}{3}\right\}+\frac{3 x}{2}-\left\{\frac{3 x}{2}\right\}+\frac{y}{2}-\left\{\frac{y}{2}\right\}+\frac{3 y}{4}-\left\{\frac{3 y}{4}\right\}=\frac{11}{6} x+\frac{5}{4} y}
\mathrm{\Rightarrow \quad\left\{\frac{x}{3}\right\}+\left\{\frac{3 x}{2}\right\}+\left\{\frac{y}{2}\right\}+\left\{\frac{3 y}{4}\right\}=0}

As each number on the L.H.S. lies in the interval \mathrm{0 \leq x< 1}. we must have

\mathrm{\left\{\frac{x}{3}\right\}=\left\{\frac{3 x}{2}\right\}=\left\{\frac{y}{2}\right\}=\left\{\frac{3 y}{4}\right\}=0}
\mathrm{\Rightarrow \quad \frac{x}{3} \cdot \frac{3 x}{2}, \frac{y}{2}\, and \, \frac{3 y}{4}}  must be integers.

\mathrm{\therefore \quad x=6,12,18,24, \quad y=4,8,12,16,20,24,28}

\mathrm{\Rightarrow } Number of ordered pairs \mathrm{(x, y) } equals \mathrm{4 \times 7=28 }.

Posted by

Irshad Anwar

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