the productive of any three consecutive natural numbers is always divisible by
Answers
Answer:
6 is the correct answer okkkkkkkkkkay
Answer:
The product of the given triplet is divisible by 6.
Step-by-step explanation:
In the given question, we have to prove that the product of any three consecutive numbers is divisible by 6
. If a number is divisible by 6
, then it means that it is also divisible by 2
and 3
.
So, let us prove that the product of any three consecutive numbers is divisible by 2
and 3
.
Consider the three consecutive numbers to be x,(x+1),(x+2)
.
For 2
:
If x
is not divisible by 2
, then it means that x
is odd.
Hence, if x
is odd, then it is a known fact that x+1
(any odd number plus 1
) is even, hence, is divisible by 2
.
Thus, out of the three consecutive numbers, at least one of them is always divisible by 2
.
For 3
:
Now, if a number is not divisible by 3
and is divided by it, then it can leave either of only two remainder – 1
and 2
.
If x
is not divisible by 3
and when divided by 3
leaves a remainder of 1
, then x+2
is going to be divisible by 3
.
If x
is not divisible by 3
and when divided by 3
leaves a remainder of 2
, then x+1
is going to be divisible by 3
.
Similarly, for the other two cases – (x+1)
and (x+2)
not being divisible by 3
, we can have that either of the other two are going to be divisible by 3
.
Thus, out of the three consecutive numbers, exactly one of them is always divisible by 3
.
Now, we have shown that out of the three consecutive numbers, one of them is always divisible by 2
and 3
. Hence, we showed that the product of any three consecutive numbers is always divisible by 6
.
Now, let us show that using some examples:
Let the triplet be 13,14,15
.
Product of the three numbers is 13×14×15=2730
.
Now, 2730÷6=455
Hence, the product of the given triplet is divisible by 6
.
Let the triplet be 16,17,18
.
Product of the three numbers is 16×17×18=4896
.
Now, 4896÷6=816
Hence, the product of the given triplet is divisible by 6