Using each of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once, how can you arrange them as a numerator and a denominator to form a fraction that is equal to 1/3?

Introduction / Context:
This puzzle combines number sense, fractions and logical reasoning. You are asked to use each of the digits from 1 to 9 exactly once to form a numerator and a denominator such that the resulting fraction is equal to one third. Problems like this train you to think systematically about digits, divisibility and equivalent fractions, which is very helpful for both competitive exams and brain-teaser style interviews.

Given Data / Assumptions:

• Available digits are 1, 2, 3, 4, 5, 6, 7, 8 and 9.
• Each digit must be used exactly once in the numerator and denominator together.
• The resulting fraction must be exactly equal to 1/3.
• Leading zeros are not allowed and the fraction should be written in standard numerator/denominator form.

Concept / Approach:
If a fraction N/D is equal to 1/3, then the denominator must satisfy D = 3 * N. Therefore, our task is to find a numerator N made from some of the digits 1 to 9 and a denominator D made from the remaining digits so that D is exactly three times N and all digits from 1 to 9 are used once. This is essentially a search problem with a strong arithmetic constraint. We can reduce the search space by using divisibility rules and length considerations for the two numbers.

Step-by-Step Solution:
Step 1: Let the fraction be N/D, with N using some digits from 1 to 9 and D using the remaining digits. Step 2: Because N/D = 1/3, we must have D = 3 * N. Step 3: We know that D must have more digits than N, because multiplying a positive integer by 3 usually increases its magnitude. A natural guess is that N is a four digit number and D is a five digit number, which together use all nine digits. Step 4: Systematically test possible four digit numerators made from 1 to 9, and for each N compute D = 3 * N. Check whether the resulting D uses exactly the remaining digits without repetition. Step 5: One such successful pair is N = 5823 and D = 17469. If you multiply 5823 by 3, you get 17469, and together the digits 5, 8, 2, 3, 1, 7, 4, 6, 9 are exactly the complete set from 1 to 9, each used once. Step 6: Therefore, the required fraction that satisfies all the conditions is 5823/17469, which equals 1/3.

Verification / Alternative check:
To verify, compute 5823 * 3. The product is 17469, which matches the denominator exactly, confirming that 5823/17469 = 1/3. You can also check that there is no repetition of digits: the numerator uses 5, 8, 2, 3 and the denominator uses 1, 7, 4, 6, 9. Combining them produces all digits 1 to 9 with no omissions or repeats. None of the other options, such as 17469/5823, give a value of one third. They yield fractions greater than 1 or with completely different values.

Why Other Options Are Wrong:
Option 17469/5823 is the reciprocal and equals 3, not 1/3. Option 2835/16947 does not satisfy the relation D = 3 * N, so its numerical value is not 1/3. Option 2583/17469 also fails the exact factor of 3 requirement and does not use digit placement correctly. Therefore these distractors do not meet the condition of forming a fraction exactly equal to one third.

Common Pitfalls:
Many learners try to guess by writing simple fractions like 123/369 and do not notice that digits repeat or that not all digits from 1 to 9 are used. Another common error is to check approximate decimal values instead of enforcing the exact relationship D = 3 * N. Testing random combinations without a strategy is time consuming and can lead to frustration. A more systematic approach, based on the algebraic condition and careful tracking of digits, is far more effective for such puzzles.

Final Answer:
The fraction using each digit 1 to 9 exactly once that equals one third is 5823/17469.