Question
Let ‘x’ be the least number divisible by 18, 20, 27,
30, and 50, and also ‘x’ is a perfect square. Find the remainder when ‘x’ is divided by 77.Solution
ATQ,
Prime factors
18 = 2 × 3²
20 = 2² × 5
27 = 3³
30 = 2 × 3 × 5
50 = 2 × 5²
LCM (take highest powers)
LCM = 2² × 3³ × 5² = 2700
Make it a perfect square
2700 = 2² × 3³ × 5² → exponent of 3 is odd
Multiply by 3 to make exponent of 3 even:
x = 2700 × 3 = 2² × 3⁴ × 5² = 8100
Remainder when x is divided by 77
8100 ÷ 77 → 77 × 105 = 8085
8100 − 8085 = 15
Remainder = 15
Statements: P = W > Y; U < W = X; T ≤ G ≤ Y
Conclusions:
I. T < P
II. P = X
III. U > G
Statements: G > N > P = E ≥ H < L; M < E < B < C = Q > X; U > W > Y = Q > H
Conclusions:
I). U > P
II). Y > P
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Conclusion: I. O > G II. W < N
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