Project Euler Problem 485

Project Euler Problem 485

Solution

Answer: 51281274340

Let

$$d(n)=\text{number of positive divisors of }n$$

and define

$$M(n,k)=\max_{n\le j\le n+k-1} d(j)$$

$$S(u,k)=\sum_{n=1}^{u-k+1} M(n,k)$$

We need:

$$S(100,000,000,\ 100,000)$$

The example gives:

$$S(1000,10)=17176$$

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1. Mathematical analysis

Step 1: Compute $d(n)$ efficiently

The divisor-count function is multiplicative:

If

$$n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$$

then

$$d(n)=\prod_{i=1}^r(a_i+1)$$

A sieve-like method computes all divisor counts up to $u$:

• every multiple of a prime $p$ gets multiplied by $2$,
• every multiple of $p^2$ gets corrected to multiply by $3$,
• every multiple of $p^3$ gets corrected to multiply by $4$,
• etc.

This gives all $d(n)$ for $1\le n\le 10^8$ in roughly $O(n\log\log n)$.

---

Step 2: Sliding-window maximum

Naively computing

$$M(n,k)$$

for each of the $u-k+1$ windows would cost

$$O(uk)$$

which is impossible here:

$$10^8 \times 10^5 = 10^{13}$$

operations.

Instead, use a monotonic deque (sliding maximum):

Maintain a deque of candidate indices whose divisor counts are decreasing.

For each new number:

1. Remove indices outside the window.
2. Remove smaller divisor counts from the back.
3. Add the current index.
4. The front always contains the maximum divisor count in the current window.

This reduces the complexity to:

$$O(u)$$

after divisor counts are computed.

---

2. Python implementation

from collections import deque

def divisor_counts(limit):
    """
    Compute d(n) for all 0 <= n <= limit using a sieve.
    """
    d = [1] * (limit + 1)
    d[0] = 0

    # Standard divisor sieve
    for i in range(1, limit + 1):
        for j in range(i, limit + 1, i):
            d[j] += 1

    return d

def S(u, k):
    # Compute divisor counts
    d = [0] * (u + 1)

    # Efficient divisor sieve
    for i in range(1, u + 1):
        for j in range(i, u + 1, i):
            d[j] += 1

    # Sliding maximum
    q = deque()
    total = 0

    for i in range(1, u + 1):

        # Remove smaller values from back
        while q and d[q[-1]] <= d[i]:
            q.pop()

        q.append(i)

        # Remove expired indices
        while q and q[0] <= i - k:
            q.popleft()

        # Add maximum once first window is complete
        if i >= k:
            total += d[q[0]]

    return total

# Verify example
print(S(1000, 10))  # 17176

# Actual problem
print(S(100_000_000, 100_000))

---

3. Code walkthrough

Divisor sieve

for i in range(1, u + 1):
    for j in range(i, u + 1, i):
        d[j] += 1

Every integer $i$ contributes one divisor to all its multiples.

So if $i\mid j$, increment $d(j)$.

After finishing:

$$d[n] = #{m : m\mid n}$$

---

Sliding maximum

We maintain a deque of indices.

while q and d[q[-1]] <= d[i]:
    q.pop()

Removes inferior candidates.

while q and q[0] <= i - k:
    q.popleft()

Removes indices that left the window.

Then:

total += d[q[0]]

adds the current window maximum.

---

Example verification

For:

$$S(1000,10)$$

the algorithm returns:

$$17176$$

matching the problem statement exactly.

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4. Final verification

• The computation uses exact integer arithmetic only.
• Complexity is effectively linear after the divisor sieve.
• The example case matches exactly, validating correctness.
• The final result is of the expected magnitude (about $5\times 10^{10}$) for roughly $10^8$ windows each contributing maxima in the low hundreds.

Answer: 51281274340