Question 65

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?

Solution

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 = 23

=> Ans - (D)

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