Question

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

Answer 1

Hint: The rule for divisibility by 6 is that the number should be divisible by 2 and 3. For 8, the rule is that the last three digits should be divisible by 8. For divisibility by 12, the number should be divisible by 3 and 4.

Complete step-by-step solution -
To find the numbers which are divisible by 6, 8 and 12, the number should also be divisible by the LCM of these three numbers. So, we will first find the LCM of 6, 8 and 12. This is given by-
$\begin{gathered}
  2\left| \!{\underline {\,
  {6,8,12} \,}} \right. \\
  2\left| \!{\underline {\,
  {3,4,6} \,}} \right. \\
  2\left| \!{\underline {\,
  {3,2,3} \,}} \right. l \\
  3\left| \!{\underline {\,
  {3,1,3} \,}} \right. \\
  1\left| \!{\underline {\,
  {1,1,1} \,}} \right. \\
\end{gathered} $
The LCM will be the product of the prime factors obtained.
$LCM\left( {6,\;8,\;12} \right) = 2 \times 2 \times 2 \times 3 = 24$
We will now find the multiples of 24. The first multiple which comes out to be of 3 digits will be our final answer.
$\begin{gathered}
  24 \times 1 = 24 \\
  24 \times 2 = 48 \\
  24 \times 3 = 72 \\
  24 \times 4 = 96 \\
  24 \times 5 = 120 \\
\end{gathered} $
Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.
This is the required answer.

Note: In such types of questions it is always advisable to divide the final number by the given numbers to ensure that they are divisible by the final answer. So,
$
  \dfrac{{120}}{6} = 20l \\
  \dfrac{{120}}{8} = 15\\
  \dfrac{{120}}{{12}} = 10 \\
$
The answer is now verified.