Question
Let βyβ be the smallest number divisible by 12, 20,
28, and 42, such that βyβ is a perfect square. What will be the remainder when βyβ is divided by 97?Solution
Prime factors 12 = 2Β² Γ 3 20 = 2Β² Γ 5 28 = 2Β² Γ 7 42 = 2 Γ 3 Γ 7 LCM LCM = 2Β² Γ 3 Γ 5 Γ 7 = 420 Make it a perfect square 420 = 2Β² Γ 3ΒΉ Γ 5ΒΉ Γ 7ΒΉ β need to multiply by 3 Γ 5 Γ 7 = 105 β y = 420 Γ 105 = 44100 = 210Β² Find remainder when divided by 97 210 β‘ 16 (mod 97) So y = 210Β² β‘ 16Β² = 256 β‘ 62 (mod 97) Final Answer: 62
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