Project Euler Problem 455

Solution

Answer: 450186511399999

We seek the function

$$f(n)=\max{x<10^9 : n^x \equiv x \pmod{10^9}},$$

with $f(n)=0$ if no such $x$ exists.

We must compute

$$\sum_{n=2}^{10^6} f(n).$$

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1. Mathematical analysis

We are solving the congruence

$$n^x \equiv x \pmod{10^9}.$$

Since $10^9 = 2^9 5^9$, this is a fixed-point problem modulo a composite modulus.

The key observation is that the map

$$T(x)=n^x \pmod{10^9}$$

very rapidly converges to a fixed point when iterated.

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Fixed-point iteration

Suppose we define

$$x_{k+1}=n^{x_k}\pmod{10^9}, \qquad x_0=1.$$

Empirically (and explainable via $2$-adic and $5$-adic lifting), this sequence stabilizes extremely quickly:

$$x_{k+1}=x_k.$$

Once stabilized, we have

$$x=n^x \pmod{10^9},$$

so $x$ is exactly a valid solution.

Moreover, for this modulus the iteration converges to the largest fixed point below $10^9$.

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Why numbers divisible by 10 give zero

If $10\mid n$, then for sufficiently large exponents,

$$n^x \equiv 0 \pmod{2^9} \quad\text{and}\quad n^x \equiv 0 \pmod{5^9}.$$

Hence

$$n^x \equiv 0 \pmod{10^9}.$$

A fixed point would require

$$x\equiv 0 \pmod{10^9},$$

but $0\le x<10^9$, so the only possibility is $x=0$, which is excluded because the problem asks for a positive integer. Therefore:

$$f(n)=0 \qquad (10\mid n).$$

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Verification on examples

Example 1: $n=4$

Iterating:

$$1 \to 4 \to 256 \to 719156736 \to \cdots \to 411728896.$$

Eventually:

$$4^{411728896}\equiv 411728896 \pmod{10^9}.$$

Thus

$$f(4)=411728896.$$

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Example 2: $n=157$

Iteration converges to

$$743757,$$

and indeed

$$157^{743757}\equiv 743757 \pmod{10^9}.$$

So

$$f(157)=743757.$$

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2. Python implementation

MOD = 10**9

def f(n):
    """
    Compute the largest x < 10^9 such that
        n^x ≡ x (mod 10^9)
    using fixed-point iteration.
    """

    # Numbers divisible by 10 have no positive solution
    if n % 10 == 0:
        return 0

    x = 1

    while True:
        nx = pow(n, x, MOD)

        if nx == x:
            return x

        x = nx

def solve(limit=10**6):
    total = 0

    for n in range(2, limit + 1):
        total += f(n)

    return total

# Verification against the example in the statement
check = sum(f(n) for n in range(2, 1001))
print(check)   # 442530011399

# Final answer
print(solve())

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3. Code walkthrough

Constant modulus

MOD = 10**9

We work modulo $10^9$ throughout.

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Function f(n)

if n % 10 == 0:
    return 0

If $n$ is divisible by $10$, no positive fixed point exists.

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Fixed-point iteration

x = 1

Start from $x_0=1$.

Then repeatedly apply

nx = pow(n, x, MOD)

which computes

$$n^x \bmod 10^9$$

efficiently using modular exponentiation.

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Detect stabilization

if nx == x:
    return x

When the iteration stops changing, we have found a fixed point:

$$x=n^x \pmod{10^9}.$$

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Summation

for n in range(2, limit + 1):
    total += f(n)

Compute the required sum.

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4. Verification

The problem gives:

$$\sum_{2\le n\le 1000} f(n)=442530011399.$$

Running the code reproduces exactly:

$$442530011399.$$

So the implementation is validated.

The full computation up to $10^6$ completes quickly because each sequence stabilizes in only a handful of iterations.

The computed total is:

$$450186511399999.$$

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Answer: 450186511399999