Question

find the out of the smallest number which is divisible by 6, 12, 18 option A 360 B 180
C 120 D 60

Answer 1

Given,

The divisors are = 6,12 and 18

Given choices of number = 360,180,120 and 60

To find,

The smallest number from given choices which can be divided by the given divisors.

Solution,

We can easily solve this mathematical problem by dividing the given numbers with the given divisors. If there was no given choices of numbers, then we can simply do the LCM or Least Common Multiple of the given divisor to find our necessary number.

First number = 60

60÷6 = 10 (Divisible)

60÷12 = 5 (Divisible)

60÷18 = Not Divisible

Second number = 120

120÷6 = 20 (Divisible)

120÷12 = 10 (Divisible)

120÷18 = Not Divisible

Third number = 180

180÷6 = 30 (Divisible)

180÷12 = 15 (Divisible)

180÷18 = 10 (Divisible)

Fourth number = 360

360÷6 = 60 (Divisible)

360÷12 = 12 (Divisible)

360÷18 = 20 (Divisible)

Now, 180 and 360 are divisible by the given divisors. But, we need the smallest number, so the necessary number will be 180.

Hence, 180 is divisible by 6,12,18. (Option C)

Answer 2

$\underline{\textsf{Given:}}$

$\textsf{Numbers are 6,12,18}$

$\underline{\textsf{To find:}}$

$\textsf{The smallest number which is divisible by ,6,12,18}$

$\underline{\textsf{Solution:}}$

$\textsf{It is enough to find the least common multiple}$

$\textsf{of 6,12 and 18}$

$\mathsf{6=2{\times}3}$

$\mathsf{12=2^2{\times}3}$

$\mathsf{18=2{\times}3^2}$

$\textsf{Choosing all the factors which having highest degree}$

$\textsf{L.C.M}\mathsf{=2^2{\times}3^2=36}$

$\implies\textsf{36 is the smallest number which is}$

$\textsf{divisible by 6,12 and 18}$

$\textsf{But 36 is not in the option}$

$\textsf{So we have to choose the smallest number in the options}$

$\textsf{which is divisible by 6,12,18}$

$\therefore\textsf{180 is the smallest in the given option which is divisible by 6,12,18}$

$\textsf{Option (B) is correct}$