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Question Number 30956 by Tinkutara last updated on 01/Mar/18

Answered by MJS last updated on 01/Mar/18

$$\left(3\right)\mathrm{because}n!\mid \underset{k=i}{\prod ^{i+n-1}}k;i\in \mathbb{N}$$

Commented by Tinkutara last updated on 01/Mar/18

But it is not greatest. The greatest number is the product itself.

Commented by MJS last updated on 01/Mar/18

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{exactly}\mathrm{one}\mathrm{of}n\mathrm{consecutive}$$$$\mathrm{numbers}m\in \mathbb{N}\mathrm{with}n\mid m,\mathrm{and}\mathrm{at}$$$$\mathrm{least}\mathrm{one}\mathrm{for}\mathrm{each}n-j,1\leqq j\leqq n-1with$$$$(n-j)\mid m$$

Commented by MJS last updated on 01/Mar/18

$$\mathrm{true},\mathrm{but}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{numbers}$$$${\mathrm{it}}^{\prime }\mathrm{s}\left(3\right)$$

Commented by Tinkutara last updated on 01/Mar/18

Yes when we see the options. Thanks got it. ��

Commented by MJS last updated on 01/Mar/18

$$...{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{always}\mathrm{the}\mathrm{question}:$$$$\mathrm{do}\mathrm{they}\mathrm{want}\mathrm{us}\mathrm{to}\mathrm{choose}\mathrm{or}$$$$\mathrm{do}\mathrm{they}\mathrm{want}\mathrm{us}\mathrm{to}\mathrm{think}?$$

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