Let [x] denote the greatest integer ≤x. Then the number of ordered pair (x,y), where x and y are positive integers less than 30 such that [(x/2)]+[((2x)/3)]+[(y/4)]+[((4y)/5)]=((7x)/6)+((21y)/(20)) is (A) 1 (B) 2 (C) 3 (D) 4


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Question Number 118930 by Ar Brandon last updated on 20/Oct/20

$Let [x] denote the greatest integer ⩽ x . Then the$ $number of ordered pair (x, y), where x and y are$ $positive integers less than 30 such that$ $[\frac{x}{2}] + [\frac{2 x}{3}] + [\frac{y}{4}] + [\frac{4 y}{5}] = \frac{7 x}{6} + \frac{21 y}{20}$ $is$ $(A) 1 (B) 2 (C) 3 (D) 4$
	Answered by floor(10²Eta[1]) last updated on 20/Oct/20

	$\frac{7 x}{6} + \frac{21 y}{20} \in N$ $\Rightarrow \frac{70 x + 63 y}{60} \in N$ $\Rightarrow 70 x + 63 y \equiv 10 x + 3 y \equiv 0 (\mod 60)$ $60 ∣ 10 x + 3 y$ $\Rightarrow 390 > 10 x + 3 y ⩾ 60$ $10 x + 3 y \in {60, 120, 180, 240, 300, 360}$ $10 x + 3 y = 60 z, 1 ⩽ z ⩽ 6$ $the solution to this diophantine equation is :$ $(x, y) = (6 z - 3 k, 10 k), k \in Z$ $\Rightarrow 0 < 10 k < 30 \Rightarrow k = 1 or k = 2$ $\Rightarrow y = 10$ $x \in {3, 9, 15, 21, 27}$ $y = 20$ $x \in {6, 12, 18, 24}$ $after checking that solutions on the$ $original equation the only one that works$ $are when y = 20 \Rightarrow x \in {6, 12, 18, 24}$ $Ans : D$
	Commented by Ar Brandon last updated on 21/Oct/20
	Thank you Sir
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