Answer:
Now we take the product of three consecutive numbers as given below. \[1\times 2\times 3=6\] \[3\times 4\times 5=60\] \[8\times 9\times 10=720\] \[17\times 18\times 19=5814\] In all the products, we have 6 or 4 or 0 in the units place, so, each product is divisible by 2. Also, sum of these all digits is divisible by 3. So every product is divisible by 3. Since 2 and 3 are co-prime. So the product \[2\times 3=6\] divides every product. Thus, the product of three consecutive numbers is always divisible by 6.
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