Number System MCQs with Answers Exercise I

By using the above formula..

no of trailing zero=(ⁿ⁄₅)+(ⁿ⁄₂₅)+(ⁿ⁄₁₂₅)+(ⁿ⁄₆₂₅)+.......................

no of trailing zero in 100!=(¹⁰⁰⁄₅)+(¹⁰⁰⁄₂₅)+(¹⁰⁰⁄₁₂₅).

no of trailing zero in 100!=20+4+0.
no of trailing zero in 100!=24.

The unit digit of power of 13 is repeated 3,9,7,1.
  so as we see the power is 45 so if we repeat 3,9,7,1 so at 45th place, the unit digit is 3.
  or
  This is for finding the unit digit every times this 3,9,7,1 four number are repeated,so
  45%4 so we will get the reminder is 1 so for 1th place 3 present in 3,9,7,1 numbers so 3 is the unit digit.

The correct answer is d) A = 5, B = 3.
For a number to be divisible by 8, the last 3 digits must be divisible by 8.
In this case, the last 3 digits are 8B2.
The only values of B that make 8B2 divisible by 8 are 0, 4, and 8.
For a number to be divisible by 11, the difference between the sum of the digits at the odd and even places must be divisible by 11.
In this case, the sum of the digits at the odd places is 4 + 3 + 6 + 8 + B = 21 + B.
The sum of the digits at the even places is 0 + 7 + 6 + 2 = 15.
Therefore, the difference between the sum of the digits at the odd and even places is 21 + B - 15 = 6 + B.
The only values of B that make 6 + B divisible by 11 are 3 and 8.
Therefore, the possible values of A and B are A = 5, B = 3 or A = 5, B = 8.
However, A = 5, B = 3 is the smallest possible values of A and B.