Question
If β48b297a54β is a nine digit number which is
divisible by β11β, then find the smallest value of (a + b).Solution
Divisibility rule of 11: (sum of the digits at odd place β sum of the digits at even place) must be multiple of 11 or equal to zero. So, (4 + b + 9 + a + 4) β (8 + 2 + 7 + 5) = 0 or a multiple of β11β. (17 + a + b) β 22 = 0 (because we want smallest value of a + b) Or, (a + b) = 22 β 17 = 5
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