Question
The product of three consecutive numbers is always divisible by $6$. Verify this statement with the help of some examples.
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Answer
Let the three consecutive numbers be n, $(n + 1), ( n + 2).$
The product of these numbers be $n(n + 1)(n + 2).$
We know that the product of three consecutive numbers is always divisible by $3.$
Out of the three consecutive numbers, one will be even. Which means the product is also divisible by $2.$
Hence the product of three consecutive numbers is divisible by $3$ and $2.$
Therefore the product of three consecutive numbers is also divisible by $6.$
$Ex.1$: Take three consecutive number $21, 22$ and $23.$
$21$ is divisible by $3.$
$22$ is divisible by $2.$
$ \therefore 21 \times 22$ is divisible by $3 \times 2 (= 6)$
$ \therefore 21 \times 22 \times 23$ is divisible by $6.$
$Ex.2$: Take three consecutive numbers $47, 48$ and $49.$
$48$ is divisible by $2$ and $3$ both.
$\therefore 48$ is divisible by $2 \times 3 = 6$
$ \therefore 47 \times 48 \times 49$ is divisible by $6.$
The product of these numbers be $n(n + 1)(n + 2).$
We know that the product of three consecutive numbers is always divisible by $3.$
Out of the three consecutive numbers, one will be even. Which means the product is also divisible by $2.$
Hence the product of three consecutive numbers is divisible by $3$ and $2.$
Therefore the product of three consecutive numbers is also divisible by $6.$
$Ex.1$: Take three consecutive number $21, 22$ and $23.$
$21$ is divisible by $3.$
$22$ is divisible by $2.$
$ \therefore 21 \times 22$ is divisible by $3 \times 2 (= 6)$
$ \therefore 21 \times 22 \times 23$ is divisible by $6.$
$Ex.2$: Take three consecutive numbers $47, 48$ and $49.$
$48$ is divisible by $2$ and $3$ both.
$\therefore 48$ is divisible by $2 \times 3 = 6$
$ \therefore 47 \times 48 \times 49$ is divisible by $6.$
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