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
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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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