Question
Find the smallest number which when divided by 7, 9 and
12 leaves a remainder 5 in each case.Solution
Let the number be N. N β‘ 5 (mod 7), (mod 9), (mod 12) β N β 5 is divisible by 7, 9 and 12 LCM(7, 9, 12): 7 = 7 9 = 3Β² 12 = 2Β² Γ 3 LCM = 2Β² Γ 3Β² Γ 7 = 4 Γ 9 Γ 7 = 252 So N β 5 = 252 β N = 252 + 5 = 257
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