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Find the least number required to be added to 3105 so that it is exactly divisible by 3, 4, 5 and 6.
- A15
- B12
- C25
- D10
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CCommunity
LCM of 3, 4, 5 and 6 = 2 × 3 × 2 × 5 × 1 × 1 = 60
On dividing 60 with 3105, we get remainder 45
∴ Number to be added = 60 - 45 = 15
Find the least number required to be added to 3105 so that it is exactly divisible by 3, 4, 5 and 6.
On dividing 60 with 3105, we get remainder 45
∴ Number to be added = 60 - 45 = 15
Find the least number required to be added to 3105 so that it is exactly divisible by 3, 4, 5 and 6.