Question
Find the least perfect square number divisible by 8, 18,
27, and 40. Then, calculate the remainder when it is divided by 121.Solution
ATQ,
LCM of 8, 18, 27 and 40 = 2160
Since, 2160 = (2 × 2 × 2 × 2 × 3 × 3 × 3 × 5)
To make it a perfect square, multiply by (5) = 5
So, required number = 2160 × 5 = 10800
10800 ÷ 121 = 89 × 121 + 81
So, desired remainder = 81
? = (597.98 ÷ (6.97 2.01 – 3.1)) × 12.9
1153, 1206, ?, 1326, 1393, 1464
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