Question-30956

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

Answered by MJS last updated on 01/Mar/18

(3) because n! ∣ \underset{k = i}{\prod^{i + n - 1}} k; i \in 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

{there}^{'} s exactly one of n consecutive

numbers m \in N with n ∣ m, and at

least one for each n - j, 1 ≦ j ≦ n - 1 w i t h

(n - j) ∣ m

Commented by MJS last updated on 01/Mar/18

true, but of the given numbers

{it}^{'} s (3)

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

\dots {it}^{'} s always the question :

do they want us to choose or

do they want us to think ?

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