Q.

The greatest positive integer, which divides (n+2)(n+3)(n+4)(n+5)(n+6) for all n∈N, is

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a

4

b

120

c

240

d

24

answer is B.

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Detailed Solution

Let P(n):(n+2)(n+3)(n+4)(n+5)(n+6) Put n=1P(1)=(1+2)(1+3)(1+4)(1+5)(1+6)=3×4×5×6×7=120×21Put n=2P(2)=(2+2)(2+3)(2+4)(2+5)(2+6)=4×5×6×7×8=120×56Hence, P(n) is always divisible by 120.
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