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The smallest 3-digit natural number that is completely divisible by 8 and 12 is:
144
108
124
120
120
Find the LCM of 8 and 12, which is 24. Simply find the smallest multiple of 24 that is greater than or equal to 100.
We are given two divisors, 8 and 12. We need to find the smallest 3-digit natural number divisible by both.
LCM(a,b)=HCF(a,b)a×b
Find the LCM of 8 and 12, which is 24. Simply find the smallest multiple of 24 that is greater than or equal to 100.
Many students find the LCM as 24 and incorrectly choose 24 or 96 (the largest 2-digit number), forgetting that the question specifically asks for the smallest 3-digit number.
Calculate the LCM
To find the numbers divisible by both 8 and 12, we must first find their Least Common Multiple (LCM). The prime factorization is 8=23 and 12=22×3. The LCM is 23×3=24.
LCM(8,12)=24
Identify multiples of 24
The numbers divisible by both 8 and 12 are multiples of 24: 24,48,72,96,120,144,…
24×1=24,24×2=48,24×3=72,24×4=96,24×5=120
Select the smallest 3-digit number
Looking at the multiples, 96 is the largest 2-digit multiple, and 120 is the smallest 3-digit multiple of 24.
120
D is correct because 120 is the smallest 3-digit multiple of the LCM of 8 and 12, which is 24.
This concept is foundational for problems involving 'ringing bells at intervals' or 'synchronizing light flashes', where you always need to calculate the LCM of the given time periods.