$18$ is divisible by both $2$ and $3$ . It is also divisible by $2 \times 3 = 6$ . Similarly, a number is divisible by both $4$ and $6$ . Can we say that the number must also be divisible by $4 \times 6 = 24$ $?$ If not give an example to justify the answer.

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Solution

Step 1: Understanding the property of divisibility.
If a number is divisible by two co-prime numbers, then it is divisible by their product also.
The numbers $2$ and $3$ are co prime numbers, if $18$ is divisible by both $2$ and $3$ , it is also divisible by $2 \times 3 = 6$ .
Step 2: Check the divisibility of a number by $4$ and $6$ .
Consider a number $36$ .
$36$ is divisible by both $4$ and $6$ , but it is not divisible by their product $4 \times 6 = 24$ .
Also, $4$ and $6$ are not co-prime numbers.
Hence, the answer is justified.

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18 is divisible by both 2 and 3. It is also divisible by 2×3=6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4×6=24 ? If not give an example to justify the answer.

$18$ is divisible by both $2$ and $3$ . It is also divisible by $2 \times 3 = 6$ . Similarly, a number is divisible by both $4$ and $6$ . Can we say that the number must also be divisible by $4 \times 6 = 24$ $?$ If not give an example to justify the answer.