Question : Find the second smallest number which, when divided by 81 or 63, leaves a remainder of 7 in each case.
Option 1: 1246
Option 2: 1141
Option 3: 1362
Option 4: 1137
Correct Answer: 1141
Solution :
If we add the remainder with the LCM of the numbers, we will get the required number.
LCM of (81, 63) = 567
The smallest number = 567k + 7 when k = 1
The second smallest number = 567k + 7 when k = 2
Hence, the required number = 567 × 2 + 7 = 1134 + 7 = 1141
Hence, the correct answer is 1141.
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