1.

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

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.


Discussion

No Comment Found