Question

greatest 3 digit number which when added to 45 , is exactly divisible by 6,8,12 will be

Answer 1

First find the lcm of 6,8,12, which is 24 now find the greatest three digit number that is divisible by 24,

How to find it, simply divide 999 by 24, whatever the remainder you will get subtract it from 24 that is your number, which is 984.
Now our required number is 984-45=939

Answer 2

The greatest number of three digits which when added to 45 is exactly divisible by 6 , 8 , 12 is 939

Given :

The numbers 6 , 8 , 12

To find :

The greatest number of three digits which when added to 45 is exactly divisible by 6 , 8 , 12

Solution :

Step 1 of 3 :

Find LCM of 6 , 8 , 12

6 = 2 × 3

8 = 2 × 2 × 2

12 = 2 × 2 × 3

∴ LCM of 6 , 8 , 12

= 2 × 2 × 2 × 3

= 24

Step 2 of 3 :

Find greatest number of three digits which is exactly divisible by 6 , 8 , 12

The greatest number of three digits = 999

When 999 is divided by 24 we get 41 as quotient and 15 as remainder

∴ The greatest number of three digits which is exactly divisible by 6 , 8 , 12

= 999 - 15

= 984

Step 3 of 3 :

Find the greatest number of three digits which when added to 45 is exactly divisible by 6 , 8 , 12

Since we have to find the greatest number of three digits which when added to 45 is exactly divisible by 6 , 8 , 12

Hence the required number

= 984 - 45

= 939

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