The greatest positive integer. which divides (n+16)(n+17)(n+18)(n+19), for all n ∈ N, is-

Question

The greatest positive integer. which divides (n+16)(n+17)(n+18)(n+19), for all n ∈ N, is- (A) 2 (B) 4 (C) 24 (D) 120

Tardigrade Admin · Accepted Answer

Since product of any r consecutive integers (क्रमागत पूर्णांक) is divisible by r ! and not divisible by r +1 ! So given product of 4 consecutive integers is divisible by 4 ! or 24 .

Q. The greatest positive integer. which divides $(n + 16) (n + 17) (n + 18) (n + 19)$ , for all $n \in N$ , is-

2

4

24

120

Since product of any $r$ consecutive integers (क्रमागत पूर्णांक) is divisible by $r$ ! and not divisible by $r + 1!$
So given product of 4 consecutive integers is divisible by $4!$ or 24 .

Q. The greatest positive integer. which divides (n+16)(n+17)(n+18)(n+19), for all n∈N, is-

2

4

24

120

Since product of any r consecutive integers (क्रमागत पूर्णांक) is divisible by r ! and not divisible by r+1! So given product of 4 consecutive integers is divisible by 4! or 24 .

Q. The greatest positive integer. which divides $(n + 16) (n + 17) (n + 18) (n + 19)$ , for all $n \in N$ , is-

Since product of any $r$ consecutive integers (क्रमागत पूर्णांक) is divisible by $r$ ! and not divisible by $r + 1!$
So given product of 4 consecutive integers is divisible by $4!$ or 24 .