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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 6.1). 152). 123). 254). 10 |
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Answer» 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. |
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