Find the smallest positive integer such that for every polynomial with integer coefficients and for every integer ,

is divisible by .

(That is, the -th derivative of at is always a multiple of .)

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My first thoughts

I started by writing a general polynomial:

Then,

At first, I thought about taking something extreme like , but the question asks for the smallest .
So I tried a simpler test case: let .

Then the -th derivative is

Since the problem says must be divisible by for every integer polynomial and every integer , it must also hold for this one.
That means

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Checking when becomes divisible by 2016

We can factor as:

Now, count how many times each prime appears in .

Prime	How to count	Needed power	Smallest that works
2	appears in 2, 4, 6, 8, …
3	appears in 3, 6
7	appears in 7

At ,

which contains all the required factors.
So .

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Dr. Pragel’s whiteboard explanation

Dr. Pragel showed a cleaner way to express everything.

He reminded us that

That means for any ,

Now, for a general polynomial

its 8th derivative is

We can rewrite that as

Then for any integer :

Here:

• are integers (since has integer coefficients),
• is an integer,
• is an integer.

So the entire sum is an integer multiple of .
And because

it follows that for every integer polynomial and every integer .

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Putting it all together

We’ve shown two parts:

1. Necessity:
   If , then is not divisible by .
   So smaller fail.

2. Sufficiency:
   If , then is always divisible by .

Therefore, the smallest integer that works is:

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My reflection

I really liked this problem because it mixes combinatorics, number theory, and calculus.
At first, it looks like it’s all about derivatives, but the heart of it is really in the divisibility of factorials and the structure of integer polynomials.

Final Answer:

Additional Note

After solving it this way, I checked the official solution from the Putnam Archive, which confirmed that is correct.
The archive’s proof uses the same key insight that Dr. Pragel pointed out — connecting to to formalize why (and therefore ) divides for all integer polynomials.