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In order to be divisible by 33, a number should be divisible be both '3' and '11'. Divisibility rule of 3 = Sum of digits should be divisible by '3'. Divisibility rule of 11 = Difference between sum of digits at odd places and even places must be either '0' or a multiple of '11'. For option 'A' = 2 + 8 + 7 + 8 + 1 + 2 = 28 {Since 28 is not divisible by '3', we may discard this option.} For option 'B' = 4 + 8 + 9 + 4 + 6 + 7 = 38 {Since 38 is not divisible by '3', we may discard this option} For option 'C' = 1 + 4 + 3 + 8 + 8 + 9 = 33 {Since, 33 is divisible by 3, this is a potential answer.} Now divisibility check for 11 = (1 + 3 + 8) - (4 + 8 + 9) = 9 {Since, '9' is not divisible by '11', we may discard this option} For option 'D' = 1 + 4 + 9 + 3 + 5 + 8 = 30 {Since, 33 is divisible by 3, this is a potential answer.} Now divisibility check for 11 = (1 + 9 + 5) - (4 + 3 + 8) = 0 So, option 'D' is divisible by 11.
Find the smallest number which when divided by 3, 4, 5 and 6 gives remainder ‘2’ in each case?
HCF and LCM of two positive numbers are 24 and 144. If one number is 50% more than the other number, then find the sum of both numbers
The HCF of 108 and 144 is ______
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Three numbers are in the ratio 2:3:4 and their HCF is 24. Find the numbers?
The least number which is exactly divisible by 5, 15 and 35 is
The least number which when divided by 10, 12, 15 and 20 leave zero remainder in each case and when divided by 24 leaves a remainder of 12 is:
