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Question

(B) The expression is the product of four consecutive integers: $(n+16)(n+17)(n+18)(n+19)$.Let $k = n+16$. Then the expression becomes $k(k+1)(k+2)(k+

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The expression is the product of four consecutive integers: $(n+16)(n+17)(n+18)(n+19)$.Let $k = n+16$. Then the expression becomes $k(k+1)(k+2)(k+3)$.This is equivalent to $4! 	imes \binom{k+3}{4}$.Since $\binom{k+3}{4}$ is always an integer for any positive integer $k$,the product of any $r$ consecutive integers is always divisible by $r!$.Here,$r = 4$,so the expression is divisible by $4! = 4 	imes 3 	imes 2 	imes 1 = 24$.Thus,the greatest positive integer that divides the product for all $n$ is $24$.

The greatest positive integer which divides $(n+16)(n+17)(n+18)(n+19)$,for all positive integers $n$,is

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