Question

Is it true that product of 3 consecutive natural numbers is always divisible by 6? Justify your answer?

Answer 1

Answer:

Yes .

Step-by-step explanation:

You take any 3 no.s . At least one

will be even . As all eves are

divisible by 2(0 is not possible) , hence the no. is

divisible by 2.

llly, the number will have at least 1

odd which is again divisible by 3(if we have 1 the 2*3=6).

So, the is divisible by 2 and 3 and

hence by 6. H.P.

Hope this will help you !

Good Luck

Answer 2

It's always true that the product of any three consecutive natural numbers is always divisible by 6.

Let the three consecutive natural numbers be x, x + 1 and x + 2.

So the product becomes,

x(x + 1)(x + 2) = x³ + 3x² + 2x.

With divisibility by 2, x can have two kinds of values, either x is being an odd  natural number (leaving remainder 1 on division by 2), or x is being an even natural number (leaving remainder 0 on division by 2).

Consider x as an odd natural number. So,

→  Since the product of any no. of odd numbers is also odd, x³ will be odd.

→  So will be 3x².

→  Product of an odd number and an even number is always even. So 2x is even.

Now,

→  Sum of even no. of odd numbers is always even. So x³ + 3x² becomes even.

→  Sum of any no. of even numbers is always even. So x³ + 3x² + 2x = (x³ + 3x²) + 2x becomes even.

So the product is divisible by 2.

Consider x as an even natural number. So all the terms x³, 3x² and 2x also become even. Hence so will be x³ + 3x² + 2x.

So we can say that the product of three consecutive natural numbers is always divisible by 2.

Now let's check its divisibility by 3.

With divisibility by 3, x can have three kinds of values, x being a multiple of 3, x leaving remainder 1 on division by 3, and x leaving remainder 2 on division by 3.

Consider x is a multiple of 3. So all the terms x³, 3x² and 2x become multiples of 3. Hence so will be x³ + 3x² + 2x.

Consider x leaves remainder 1 on division by 3. So,

→  x³ also leaves remainder 1 on division by 3.

→  3x² is always a multiple of 3, since it's 3 multiplied with x.

→  2x leaves remainder 2 on division by 3.

Now,

→  Since x³ and 2x leave remainders 1 and 2 respectively on division by 3, x³ + 2x becomes divisible by 3. [∵ It's true that a + b is divisible by n if a and b leave remainders k and n - k respectively on division by n.]

So, since 3x² is divisible by 3, then so will be x³ + 3x² + 2x = (x³ + 2x) + 3x².

Consider x leaves remainder 2 on division by 3. So,

→  x³ leaves remainder 2 on division by 3.

→  The remainder left by x on division by 3 does not affect 3x²! It's always divisible by 3!

→  2x leaves remainder 1 on division by 3.

As mentioned above, since x³ and 2x leave remainders 2 and 1 respectively on division by 3, x³ + 2x is divisible by 3. Hence so will be x³ + 3x² + 2x.

So we can say that the product of any three consecutive natural numbers is divisible by 3.

Divisibility rule says that a number is divisible by 6 if and only if that number is divisible by both 2 and 3.

So we came to the conclusion that the product is divisible by 6.

Done!