Question Number 30956 by Tinkutara last updated on 01/Mar/18
Answered by MJS last updated on 01/Mar/18
$$\left(\mathrm{3}\right)\:\mathrm{because}\:{n}!\mid\underset{{k}={i}} {\overset{{i}+{n}−\mathrm{1}} {\prod}}{k};\:{i}\in\mathbb{N} \\ $$
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
$$\mathrm{there}'\mathrm{s}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:{n}\:\mathrm{consecutive} \\ $$$$\mathrm{numbers}\:{m}\in\mathbb{N}\:\mathrm{with}\:{n}\mid{m},\:\mathrm{and}\:\mathrm{at} \\ $$$$\mathrm{least}\:\mathrm{one}\:\mathrm{for}\:\mathrm{each}\:{n}−{j},\:\mathrm{1}\leqq{j}\leqq{n}−\mathrm{1}\:{with} \\ $$$$\left({n}−{j}\right)\mid{m} \\ $$
Commented by MJS last updated on 01/Mar/18
$$\mathrm{true},\:\mathrm{but}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{numbers}\: \\ $$$$\mathrm{it}'\mathrm{s}\:\left(\mathrm{3}\right) \\ $$
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
$$…\mathrm{it}'\mathrm{s}\:\mathrm{always}\:\mathrm{the}\:\mathrm{question}: \\ $$$$\mathrm{do}\:\mathrm{they}\:\mathrm{want}\:\mathrm{us}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{or} \\ $$$$\mathrm{do}\:\mathrm{they}\:\mathrm{want}\:\mathrm{us}\:\mathrm{to}\:\mathrm{think}? \\ $$