Question
Let βxβ be the least number divisible by 18, 20, 27,
30, and 50, and also βxβ is a perfect square. Find the remainder when βxβ is divided by 77.Solution
ATQ,
Prime factors
18 = 2 Γ 3Β²
20 = 2Β² Γ 5
27 = 3Β³
30 = 2 Γ 3 Γ 5
50 = 2 Γ 5Β²
LCM (take highest powers)
LCM = 2Β² Γ 3Β³ Γ 5Β² = 2700
Make it a perfect square
2700 = 2Β² Γ 3Β³ Γ 5Β² β exponent of 3 is odd
Multiply by 3 to make exponent of 3 even:
x = 2700 Γ 3 = 2Β² Γ 3β΄ Γ 5Β² = 8100
Remainder when x is divided by 77
8100 Γ· 77 β 77 Γ 105 = 8085
8100 β 8085 = 15
Remainder = 15
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