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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Question : What is the smallest three-digit number which, when divided by 8 or 6, leaves a remainder of 1 in each case?
Option 1: 121
Option 2: 119
Option 3: 123
Option 4: 125
Question : What is the smallest three-digit number when divided by 2, 3, and 4 leaving the remainder 1 in each case?
Option 1: 111
Option 2: 105
Option 3: 101
Option 4: 109
Question : What is the greatest four-digit number which when divided by 3 and 4 leaves remainder 2 in each case?
Option 1: 9994
Option 2: 9996
Option 3: 9998
Option 4: 9995
Question : A number, when divided by 80, leaves a remainder of 20. What is the remainder when the same number is divided by 16?
Option 1: 2
Option 2: 4
Option 3: 6
Option 4: 8
Question : The least number which when divided by 14, 18, and 36 leaves 1 as the remainder in each case is:
Option 1: 250
Option 2: 252
Option 3: 253
Option 4: 251
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