Q. 65.0( 1 Vote )

Prove that product of four consecutive positive integers is divisible by 24.

Answer :

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

Product = n(n + 1)(n + 2)(n + 3)


We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.


For, n = 3k


Product = n(n + 1)(n + 2)(n + 3)


= 3k(3k + 1)(3k + 2)(3k + 3)


= 33k(3k + 1)(3k + 2)(k + 1)


Therefore it is divisible by 3.


For, n = 3k + 1


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)


= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)


Therefore it is divisible by 3.


For, n = 3k + 2


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)


= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)


Therefore it is divisible by 3.


n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2


For n = 4p


Product = n(n + 1)(n + 2)(n + 3)


= 4p(4p + 1)(4p + 2)(4p + 3)


= 24p(4p + 1)(2p + 1)(4p + 3)


= 8p(4p + 1)(2p + 1)(4p + 3)


Therefore it is divisible by 8.


For n = 4p + 1


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)


= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)


= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 3


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)


= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)


= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)


Therefore, it is divisible by 8.


Since it is divisible by both 3 and 8 and both are mutually prime numbers .


Therefore, it will be divisible by 38 = 24


So, product of four consecutive positive integers is divisible by 24.


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