Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.8 AMC 10/12 Find the answer South Africa

All four-digit positive integers which are rearrangements of the number $2316$ are written in increasing order. What is the largest possible difference between two adjacent numbers in this list?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

$6123$ is the smallest number starting with a $6$ and $3621$ is the largest number starting with a $3$ . The difference between these is $6123 - 3621 = 2502$ . Adjacent numbers between $1236$ and $3621$ will be less than $2500$ apart, while adjacent numbers larger than $6123$ will be less than $1000$ apart. Thus $2502$ .

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