Question
Determine the smallest perfect square that is exactly
divisible by 9, 15, 20, and 35. Then, find the remainder when this perfect square is divided by 88.Solution
ATQ,
LCM of 9, 15, 20 and 35 = 1260
Since, 1260 = (2 × 2 × 3 × 3 × 5 × 7)
To make it a perfect square, multiply by (2 × 5 × 7) = 70
So, required number = 1260 × 70 = 88200
88200 ÷ 88 = 1002 × 88 + 24
So, desired remainder = 24
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