#1000 Problem $1000$ - Project Euler

Problem $1000$<br>Published on Saturday, 6th June 2026, 08:00 pm; Solved by 3<br>Problem 1000

This problem contains three sub-problems and a meta-problem. Enter the answer to the meta-problem as final answer.

Sub-problem: Max And

The numbers $1$ to $n$ are to be divided into two groups $A$ and $B$. Let $I(n)$ be the maximum possible value of $$\sum_{a\in A}\sum_{b\in B}a\wedge b$$<br>where $\wedge$ is the bitwise AND operator.

It can be verified that $I(10)=50$, if for example $A=\{1,4,7,10\}$ and $B=\{2,3,5,6,8,9\}$, although there are a number of other solutions as well.

Find $I(1000)$.

Sub-problem: Max Xor Sum

For two integers $x, y$, write $[x, y]$ for the bitwise XOR of $x^2$ and $y^2$.

A finite sequence of integers $a_0, a_1, \dots, a_r$ satisfies the following properties:

$1 \le a_i \le N$ for all $0 \le i \le r$;

$[a_{i - 1}, a_i] \lt [a_i, a_{i + 1}]$ for all $0 \lt i \lt r$.

Let $X(N)$ be the maximal possible value of the sum $\sum_{i = 1}^r [a_{i - 1}, a_i]$.

For example, $X(4) = 71$ can be achieved with the sequence $2,1,3,2,4,3$. Also $X(10) = 702$.

Find $X(1000)$.

Sub-problem: Unreachable Nim

Two players are playing a three-pile Nim game. The game status is an ordered triple $(a, b, c)$, representing the number of stones in each pile.

A player will always make a winning move, if there is at least one; otherwise, any valid move can be made, unless there is no valid move left, at which point the game ends.

A game status is called unreachable if it never appears during the game, unless when it is the initial position. For example, the game status $(1, 1, 1)$ is unreachable.

Let $C(N)$ be the number of unreachable statuses with $0 \le a, b, c \lt N$.<br>You are given $C(10) = 123$.

Find $C(1000)$.

Meta-problem:

A sequence $M$ is defined by

$M(0) = I(1000)$;

$M(1) = X(1000)$;

$M(2) = C(1000)$;

$M(k) = M(k - 1)M(k - 2)M(k - 3)$ for $k \ge 3$.

You are given $M(4) \equiv 457587170 \pmod{10^9 + 7}$.

Find $M(1000) \bmod (10^9 + 7)$.

Hint: one may assume that the given value of $M(4)$ is correct.

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