Question

what is the largest 4 digit number and smallest 3 digit number dibvisible by 615 21 24

Answer 1

6 = 2 × 3

15 = 5 × 3

21 = 7 × 3

24 = 2 × 2 × 2 × 3

So, = 2 × 2 × 2 × 3 × 5 × 7 = 840.

The largest number in 4 digits = 9999

Now, 9999 = (840 × 11) + 759

So, remainder is = 759

So the largest number, divisible by 6, 15, 21 and 24 is

= 9999 - 759

= 9240.

The smallest number in 3 digits, divisible by 6, 15, 21 and 24 is

= lcm of 6, 15, 21 and 24

= 840.