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