Answer
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.
Question 8 Exercise 3.7
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Answer:
SOLUTION:
To find the smallest digit divisble by 6,8 and 12
we need to find the LCM of 6,8 and 12
6,8,12=2\times2\times2\times3=24
100+24-4=120
120 is the smallest digit divisible by 6,8 and 12.
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