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
If 10 A 5 B 6 C 2 = 53 and 6 A 4 B 8 C 2 = 28, then 17 A 5 B 18 C 9 =?
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