Question

Prove that product of any 3 consecutive number is divisible by 6​

Answer 1

Answer:

Prove that product of any 3 consecutive number is divisible by 6

Step-by-step explanation:

let the no. be (x) , (x + 1) ,(x + 2).

whene  a number is divided by 3, the remainder obtained is 0 or 1 or 2.

therefore,

x = 3n or (3n + 1) or (3n + 2)

if x = 3n, then x is divisible by 3

if x = 3n + 1 ,then x + 2 = 3n + 1 + 2 = 3n + 3

=> x = 3(n + 1) is divisible by 3

if x = 3n + 2, then x + 1 = 3n + 2 + 1

=> 3n + 3 = 3(n + 1)

so, we can say that one of the numbers  n, n + 1 and n + 2 is always divisible by 3.

 n (n + 1) (n + 2) is divisible by 3.

now,

similarly, when a no. is divisible by 2 remainder abtained is 0 or 1.

therefore,

x = 2r or (2r + 1)

if x = 2r, then x = 2r and (x + 2) then,2r + 2

=> 2(r + 1) are divisible by 2

if x = (2r + 1), then x + 1 = 2r + 1 + 1 = 2(r + 1)is divisible by 2.

So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.

 x (x + 1) (x + 2) is divisible by 2.

 

n (n + 1) (n + 2) is divisible by 2 and 3.

 

therefore,

 n (n + 1) (n + 2) is divisible by 6.

Answer 2

Answer:

true

Step-by-step explanation:

for a number is divisible by 6 it should be both 2&3

take any 3 consecutive numbers

we have atleast one even number

case 1:

odd x even x odd= even (divisible by 2)

case2:

even x odd x even = even (divisible by 2)

Test of divisibility of 3

let 3 numbers be

n-1,n,n+1

f

we need (n-1) x n x (n+1)

n+n-1+n+1= 3n hence it is divisible by 3

hence product of any 3 consecutive numbers

is divisible by 6