Question
Let βxβ be the least number divisible by 16, 18, 20,
25 and 45, and also βxβ is a perfect square. Find the remainder when βxβ is divided by 88.Solution
ATQ, LCM of 16, 18, 20, 25 and 45 = 7200 Since, 7200 = (2 Γ 2 Γ 2 Γ 2 Γ 3 Γ 3 Γ 5 Γ 5) To make it a perfect square, multiply by β3β. So, desired number = 7200 Γ 3 = 21600 21600 Γ· 88 = 245 Γ 88 + 20 So, desired remainder = 20
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