Let x be the least multiple of 29 which when divided by 20, 21, 22, 24 and 28 then the remainders are 13, 14, 15, 17 and 21 respectively. What is the sum of digits of x?
Since, the difference between the all the numbers are : (20-13), (21-14) = 7
Thus, L.C.M. (20,21,22,24,28) = 9240
Hence, required number will be of the form = $$9240k-7=$$ multiple of 29
By hit and trial, we see that $$k=2$$ satisfy above equation, thus required number = $$(9240\times2)-7=18473$$
=> Sum of digits =Â 23
=> Ans - (D)
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