Question
Let N be the least number which when divided by 15, 25,
35, 40, and 42, the remainder in each case is 1 and N is divisible by 271. What is the sum of digits of N?Solution
LCM of 15, 25, 35, 40, and 42 15=3×5 25 = 5 x 5 35 = 5 x 7 40 = 2 x 2 x 2 x 5 42=2 x 3 x 7 LCM = (2 x 2 x 2 x 3 x 5 x 5 x 7) = 4200 Now, So, the number is in the form of 4200N + 1 And, the number is also divisible by 271. Now, When N = 1 4200 x 1 + 1 = 4201, which is not divisible by 271 When N = 2 4200 x 2 + 1 = 8401, It is divisible by 271 Now, The sum of digits of 8401 = (8 + 4 + 0+ 1) = 13 The required sum of digits of N is 13.
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