# Find the smallest number of six digits divisible by 18, 24 and 30

To find the smallest number of six digits divisible by 18, 24 and 30 we find the l.c.m. of them.

18 = 3 × 3 × 2

24 = 2 × 2 × 2 × 3

30 = 3 × 2 × 5

Therefore, l.c.m.(18, 24, 30) = 2 × 2 × 2 × 3 × 3 × 5 = 360

We know that the smallest six digit number is 100000.

Dividing 100000 by 18 = = 277. 77

So, the integer bigger and nearest to 277. 77 is 278.

Therefore, required integer = 278 × 360 = 100080

So, the smallest number of six digits divisible by 18, 24 and 30

is 100080.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Interactive Quiz:Euclid's Division Lemma44 mins  Fundamental Theorem of Arithmetic-238 mins  Champ Quiz | Fundamental Principle Of Arithmetic41 mins  Fundamental Theorem of Arithmetic- 143 mins  NCERT | Imp. Qs. on Rational and Irrational Numbers44 mins  Euclids Division Lemma49 mins  Quiz | Imp Qs on Real Numbers37 mins  Interactive Quiz - HCF and LCM32 mins  Relation Between LCM , HCF and Numbers46 mins  Application of Euclids Division Lemma50 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 