Question
Let ‘n’ be the smallest perfect square divisible by 22,
26, 33 and 39. What is the remainder when ‘n’ is divided by 100?Solution
ATQ,
LCM of 22, 26, 33 and 39 = 2574
Since, 2574 = (2 × 3 × 11 × 13 × 3)
= 2 × 3 × 3 × 11 × 13
To make it a perfect square, multiply by (2 × 11 × 13) = 286
So, required number = 2574 × 286 = 736164
736164 ÷ 100 = 7361 × 100 + 64
So, desired remainder = 64
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