Question
The biggest three-digit number leaves 2 as a remainder when
divided by 11, 9, and 4. What is the sum of its digits?Solution
ATQ,
LCM of (11, 9, and 4) = 396
Largest 3-digit number divisible by 396 = 396 × 2 = 792
So, N = 792 + 2 = 794
Sum of digits of 'N' = 7 + 9 + 4 = 20
√(3x-2)+ √(3x+2) = 4+ 2√5 then value of x is.
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