Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

NCERT Math Class 6th Question – 4 Exercise 3.7

Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Solution:

The smallest 3-digit number = 100
Since LCM of 6, 8 and 12 is divisible by them.

∴ LCM of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24
Since, all the multiples of 24 will also be divisible by 6, 8 and 12.

So, the smallest multiple of 24 in three digits will be just above
100 = (100 – 4) + 24 = 96 + 24 = 120
Hence, the required number is 120.

NCERT Math Class 6th Question – 1 Exercise 3.7Click here
NCERT Math Class 6th Question – 2 Exercise 3.7Click here
NCERT Math Class 6th Question – 3 Exercise 3.7Click here
NCERT Math Class 6th Question – 5 Exercise 3.7Click here
NCERT Math Class 6th Question – 6 Exercise 3.7Click here
NCERT Math Class 6th Question – 7 Exercise 3.7Click here
NCERT Math Class 6th Question – 8 Exercise 3.7Click here
NCERT Math Class 6th Question – 9 Exercise 3.7Click here
NCERT Math Class 6th Question – 10 Exercise 3.7Click here
NCERT Math Class 6th Question – 11 Exercise 3.7Click here

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