Is it integer?

Probability Level 5

Let $n$ be a natural number. If $\dfrac{(n^{2}!)}{(n!)^{a}}$ is an integer for all $a = 0,1,2, \ldots,b$ . Find the maximum value of b in terms of n for which this is satisfied for every n. For example if b= n+8, then for n=5, $\dfrac{(n^{2}!)}{(n!)^{a}}$ would be integer for a=1,2,3,...,b+8 i.e. 13 .

Clarification:

Choose the value of b so that for every natural number this would be true. There can be some numbers where b=n+9 would be true(in the example) but we have to enter b=n+8 since it is valid for every natural number.

n+2

n+10

1

n-1

n^{2}

n+1

n

2

1 solution

Prince Loomba
May 6, 2016

Consider $n^{2}$ objects to be divided to n people, each getting n objects. By grouping theory, No. of ways = $\frac{(n^{2})!}{(n!)^{n} \times (n!)} = \frac{(n^{2})!}{(n!)^{n+1}}$ Thus $a=n+1$

This only shows that the largest integer is at least $n+1$ . Why can't it be $n+2$ for some $n$ ?

Calvin Lin Staff - 5 years, 1 month ago

Now I have mentioned 'general' in the question.

Prince Loomba - 5 years, 1 month ago

What does "general value" mean?

Calvin Lin Staff - 5 years, 1 month ago

@Calvin Lin – General value means related to n, for example, if n=1,b=2 and so on

Prince Loomba - 5 years, 1 month ago

@Prince Loomba – That is not a standard term. I don't think other people will understand what you mean by that.

If you are asking for "general term of a sequence", then yes that makes sense. But we don't have a sequence here.

Calvin Lin Staff - 5 years, 1 month ago

@Calvin Lin – Please change the question accordingly. This is my last try to explain the question

Prince Loomba - 5 years, 1 month ago

@Prince Loomba – Before I edit, can you answer my question in my comment? That will help me decide how best to phrase this problem.

This only shows that the largest integer is at least $n+1$ . Why can't it be $n+2$ for certain values of $n$ ?
E.g. When $n = 1$ , there is no largest value $a$ , so saying that $a = n+1$ is wrong at least for $n=1$ .

Note: I think this is an interesting problem, and worth phrasing it in a clear manner for others to understand exactly what you mean / are thinking.

Calvin Lin Staff - 5 years, 1 month ago

@Calvin Lin – I mean that for b=n+1, it is always an integer but for only some values, b=n+2 holds therefore the answer is that which is true for all values of n and that is n+1 and so the answer

Prince Loomba - 5 years, 1 month ago

@Prince Loomba – K, that makes the problem much harder to clearly express. I will have to think about how to phrase this "best function which satisfies for all values but need not be the best in every case".

Calvin Lin Staff - 5 years, 1 month ago