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What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder ?
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- 312
- 962
- 1562
- 1586
Correct Option: B
We find LCM of 3, 5, 6, 8, 10 and 12 = 120
∴ Required number = 120q + 2, which is exactly divisible by 13.
Required number = 120q + 2 = 13 × 9q + 3q + 2
Clearly 3q + 2 should be divisible by 13.
For q = 8 , 3q + 2 is divisible by 13.
∴ Required number = 120q + 2 = 120 × 8 + 2
Required number = 960 + 2 = 962
