Find the second smallest number which, when divided by 81 or 63, leaves a remainder of 7 in each case.
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 1141.
LCM of (81, 63) = 567
According to the question,
The smallest number = 567x + 7, when x = 1
The second smallest number = 567x + 7, when x = 2
Therefore required number = 567 × 2 + 7 = 1134 + 7 = 1141
Hence the second smallest number will be 1141.