To solve this problem, we aim to determine how many pairs of numbers $(a, b)$ can be formed using the digits $\{1, 2, 3, 4, 5, 6\}$, with each digit appearing exactly once across the two numbers, such that their sum equals 750. Let's analyze step by step:

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Step 1: Understand the Problem

• Tom wrote two numbers $a$ and $b$.
• The numbers $a$ and $b$ use all six digits from the set $\{1, 2, 3, 4, 5, 6\}$ exactly once between them.
• Their sum is $a + b = 750$.

Step 2: Partitioning the Digits

Since $a$ and $b$ are composed of the digits $\{1, 2, 3, 4, 5, 6\}$, each digit must appear in either $a$ or $b$ exactly once. There are $\binom{6}{3} = 20$ ways to partition the six digits into two groups of three.

For each partition, we calculate all possible pairings $(a, b)$ and check if their sum is 750.

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Step 3: Permuting the Digits

After partitioning, each group of three digits can be arranged into a number in $3! = 6$ ways. Therefore, for any specific partition:

• There are $6 \times 6 = 36$ pairings of numbers $(a, b)$.

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Step 4: Checking the Constraint

For each partition, calculate $a + b$. Retain only the pairs where $a + b = 750$.

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Step 5: Calculations

We will now iterate through all partitions, compute all possible sums, and count the valid pairs.

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Key Observations:

1. The total sum of the digits $\{1, 2, 3, 4, 5, 6\}$ is: $1 + 2 + 3 + 4 + 5 + 6 = 21$
2. If $a + b = 750$, the larger number among $a$ and $b$ must be less than 750, and the sum of digits in $a$ and $b$ must balance correctly to satisfy the sum constraint.

Let me compute the exact result.from itertools import permutations, combinations

The digits used

digits = {1, 2, 3, 4, 5, 6} target_sum = 750

Store valid pairs

valid_pairs = []

Iterate through all ways to partition the digits into two groups of 3

for group1 in combinations(digits, 3): group1_set = set(group1) group2_set = digits - group1_set # Remaining digits go to the second group

# Generate all permutations of the digits in each group to form numbers
for perm1 in permutations(group1_set):
    for perm2 in permutations(group2_set):
        num1 = int("".join(map(str, perm1)))  # First number
        num2 = int("".join(map(str, perm2)))  # Second number

        # Check if the sum equals the target
        if num1 + num2 == target_sum:
            valid_pairs.append((num1, num2))

Count unique pairs

total_valid_pairs = len(valid_pairs) total_valid_pairs88successThere are 8 possible pairs of numbers $(a, b)$ that satisfy the given conditions, where $a + b = 750$ and each digit from $\{1, 2, 3, 4, 5, 6\}$ is used exactly once across the two numbers.

Would you like to see the specific pairs or further explanations?

Related Questions:

1. What are the exact pairs of numbers $(a, b)$ that satisfy the conditions?
2. How can we generalize this problem to larger digit sets?
3. What is the mathematical significance of partitioning digits for problems like this?
4. How do constraints like $a + b = 750$ affect the number of solutions in general problems?
5. Can we automate a similar approach for solving puzzles involving digit-based constraints?

Tip: Always verify the total digit sum constraints when dealing with problems requiring unique partitions of digits.