Let $f(x)$ be a polynomial with integer coefficients. There exist distinct integers $p,$ $q,$ $r,$ $s,$ $t$ such that
\[f(p) = f(q) = f(r) = f(s) = 10\]and $f(t) > 10.$ What is the smallest possible value of $f(t)?$
Hint(s):
A lot of the values of the polynomial $f(x)$ are equal to $10.$ What polynomial could you define in terms of $f(x)$, that would have a lot of roots?

We are given that $ f(x) $ is a polynomial with integer coefficients, and there exist distinct integers $ p, q, r, s $ such that:

\[
f(p) = f(q) = f(r) = f(s) = 10,
\]
and $ f(t) > 10 $.

We are asked to find the smallest possible value of $ f(t) $.

### Step 1: Define a new polynomial
To simplify the problem, define a new polynomial $ g(x) $ in terms of $ f(x) $ by subtracting 10 from $ f(x) $:

\[
g(x) = f(x) - 10.
\]

Since $ f(p) = f(q) = f(r) = f(s) = 10 $, it follows that:

\[
g(p) = g(q) = g(r) = g(s) = 0.
\]

This means that $ p, q, r, s $ are roots of the polynomial $ g(x) $. Therefore, we can express $ g(x) $ as:

\[
g(x) = c(x - p)(x - q)(x - r)(x - s),
\]
where $ c $ is a constant, and $ c $ is an integer because $ f(x) $ has integer coefficients.

Thus, we have:

\[
f(x) = g(x) + 10 = c(x - p)(x - q)(x - r)(x - s) + 10.
\]

### Step 2: Analyze the condition $ f(t) > 10 $
We are also given that $ f(t) > 10 $. Using the expression for $ f(x) $, this translates to:

\[
f(t) = c(t - p)(t - q)(t - r)(t - s) + 10 > 10,
\]
which simplifies to:

\[
c(t - p)(t - q)(t - r)(t - s) > 0.
\]

This inequality implies that the sign of $ c(t - p)(t - q)(t - r)(t - s) $ must be positive. Therefore, $ c $ and the values $ t - p, t - q, t - r, t - s $ must be chosen so that their product is positive.

### Step 3: Minimize $ f(t) $
To minimize $ f(t) $, we should choose $ t $ as close as possible to $ p, q, r, s $. Let's assume that $ p < q < r < s $ are consecutive integers to minimize the gap between them. For simplicity, let's try $ p = 1, q = 2, r = 3, s = 4 $, which gives:

\[
f(x) = c(x - 1)(x - 2)(x - 3)(x - 4) + 10.
\]

Now, we want to choose $ t $ such that $ f(t) > 10 $. The value of $ t $ should make the product $ (t - 1)(t - 2)(t - 3)(t - 4) $ positive. Let's first compute this for $ t = 5 $:

\[
f(5) = c(5 - 1)(5 - 2)(5 - 3)(5 - 4) + 10 = c(4)(3)(2)(1) + 10 = 24c + 10.
\]

For $ f(5) > 10 $, we need $ 24c > 0 $, which implies $ c > 0 $. The smallest integer value for $ c $ is $ c = 1 $, which gives:

\[
f(5) = 24 \times 1 + 10 = 34.
\]

Thus, the smallest possible value of $ f(t) $ is $ \boxed{34} $.

### Summary
The smallest possible value of $ f(t) $ is $ 34 $, achieved when $ f(x) = (x - 1)(x - 2)(x - 3)(x - 4) + 10 $ and $ t = 5 $.

Do you have any questions or would you like further details?

### Related Questions
1. What happens if we choose different values for $ p, q, r, s $?
2. How does the value of $ c $ affect the minimum of $ f(t) $?
3. Can we achieve a lower value of $ f(t) $ by considering non-consecutive integers for $ p, q, r, s $?
4. How would the solution change if $ f(t) $ needed to be greater than a value other than 10?
5. What is the general solution for $ f(x) = c(x - p)(x - q)(x - r)(x - s) + d $?

### Tip:
Always check the constraints on the problem, like distinct integers or specific ranges, when setting up your polynomial roots!

Let $f(x)$ be a polynomial with integer coefficients. There exist distinct integers $p,$ $q,$ $r,$ $s,$ $t$ such that \[f(p) = f(q) = f(r) = f(s) = 10\] and $f(t) > 10.$ What is the smallest possible value of $f(t)?

Explore how to find the smallest possible value of a polynomial with integer coefficients when given specific conditions on its roots. Learn key polynomial concepts and techniques for solving this challenging problem, including using the Fundamental Theorem of Algebra and root manipulation.

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