Question

What are the largest 4 digit and smallest 3 digit numbers divisible by 6,15,21 & 24

Answer 1

Answer:

Step-by-step explanation:

6 = 2 × 3

15 = 5 × 3

21 = 7 × 3

24 = 2 × 2 × 2 × 3

So, lcm = 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.

Answer 2

Answer:

The largest 4 digit number is 9240 and smallest 3 digit number is 840.

Step-by-step explanation:

To find : What are the largest 4 digit and smallest 3 digit numbers divisible by 6,15,21 & 24 ?

Solution :

First we find the LCM of the numbers 6,15,21 & 24

2 | 6  15  21  24

2 | 3  15  21   12

2 | 3  15   21   6

3 | 3  15   21   3

5 | 1   5     7    1

7 |  1   1     7    1

  |  1   1     1     1

$LCM(6,15,21,24)=2\times 2\times 2\times 3\times 5\times 7$

$LCM(6,15,21,24)=840$

The largest number in 4 digits = 9999

Now, $9999 = (840\times 11) + 759$

Remainder = 759

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

$n= 9999 - 759$

$n= 9240$

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

i.e. n=840