Question
Let βyβ be the smallest number divisible by 12, 20,
28, and 42, such that βyβ is a perfect square. What will be the remainder when βyβ is divided by 97?Solution
ATQ,
LCM of 12, 20, 28 and 42 = 840
Since, 840 = (2 Γ 2 Γ 2 Γ 3 Γ 5 Γ 7)
To convert this into a perfect square, we multiply by (3 Γ 5 Γ 7) = 105
So, required number = 840 Γ 105 = 88200
88200 Γ· 97 = 909 Γ 97 + 27
So, desired remainder = 27
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