The 6174 Attractor

Use four decimal digits, allowing leading zeroes such as 0352. Sort all four digits into the largest arrangement and the smallest arrangement, then subtract the smaller from the larger. Write the result in four places, adding leading zeroes if needed, before sorting again.

For 3524, the first subtraction is 5432 − 2345 = 3087. The next is 8730 − 0378 = 8352, followed by 8532 − 2358 = 6174. At 6174 the same operation gives 7641 − 1467 = 6174, so repeating it no longer changes the value.

Every four-place start with at least two different digits reaches 6174 within seven steps. Four identical digits form the exception: 1111 − 1111 = 0000, and 0000 stays at zero. The leading-zero convention is part of the rule, not a requirement that the starting value be greater than 1000.

Two things are quietly happening.

The rule cannot see the order you typed. Its first act is to sort. So 3524, 2345 and 5432 all take the same first step, and from there their runs are identical. Ten thousand starting values behave like only 715 different ones, which is already a far smaller problem than it looked.

The rule cannot produce most numbers. Subtracting two arrangements of the same digits turns out to be very constraining: only certain values can ever appear. So after a single step, the routine is already confined to a small, structured set - and once you are inside a small set and every element has somewhere to go, you cannot wander forever. You must eventually repeat yourself.

Switch to the state space view and both facts become visible at once. Everything drains inward, no branch is longer than seven rings, and the ten values that never arrive sit off to one side.

Here is the constraint. Write the sorted digits as $a\ge b\ge c\ge d$. The largest arrangement is $1000a+100b+10c+d$ and the smallest is $1000d+100c+10b+a$. Subtract:

Both coefficients are multiples of nine. Every value after the first step is divisible by nine - which immediately rules out eight of every nine numbers.

The same argument works in any base $b$. The two arrangements are permutations of one another, so they have the same digit sum, so they are congruent modulo $b-1$, so their difference is divisible by $b-1$. Nothing about base ten is special here.

Let $K$ be the map sending a four-digit decimal string to the difference between the descending and ascending arrangements of its digits, with the result written back as four digits (retaining leading zeroes).

Then for every four-digit string $n$ whose digits are not all equal:

and $K(6174)=6174$. For the ten strings with all digits equal, $K(n)=0000$ and $K(0000)=0000$.

Note what the statement is careful about: it is about four-digit strings, not about integers below ten thousand. Those are different objects, and the difference matters - 0999 is a legitimate state of this system and 999 is not.

The proof is exhaustion, and unusually, that is not a compromise.

There are exactly 10,000 four-digit strings. Each has exactly one successor. So the whole system is a finite directed graph in which every node has out-degree one - and such a graph is always a set of cycles with trees feeding into them, with no third possibility.

Enumerating all 10,000 settles it completely: there are exactly two cycles, both of length one

• 6174 and 0000 - and every valid start reaches 6174 within seven steps. The state space view is not an illustration of this argument. It is the argument, drawn.

One trap worth naming, because it is easy to fall into and this page fell into it first. Since the routine sorts its input, it is tempting to treat the digit multiset as the state and work with 715 nodes instead of 10,000. That is right about transitions and wrong about states. 6174 and 4176 have the same digits, but 6174 is the fixed point while 4176 still needs one more step - and 4176 really does occur: any start whose sorted digits satisfy $a-d=4$ and $b-c=2$ produces it, such as 9755. Collapsing the two together understates the distance of every path that lands on a rearrangement of 6174 rather than on 6174 itself. The distances here are measured on numbers for that reason.

This is worth dwelling on, because it is the exception rather than the rule. The computation here is a proof only because the space is finite and every case was checked. The same computation over an infinite space would prove nothing whatsoever, however many cases it got through - which is exactly the situation for the Collatz map, whose statement looks almost identical to this one and remains open after eighty years.

Three honest boundaries.

The constant is not a property of the rule. It is a property of the rule together with base ten and four digits. Three digits give 495. Most other combinations of base and width give cycles of length greater than one, or several attractors, or both. Nothing above explains why four decimal digits should be so unusually tidy - it is checked, not understood.

Divisibility by nine is not the whole explanation. It shows the space collapses. It does not single out 6174, and no short argument here does. If you want the destination rather than the shrinkage, the enumeration is currently the honest answer.

Nor is "sorting means the multiset is the state". As above, that shortcut is false in exactly one place - at the attractor - and the failure is invisible unless you check. It is a fair miniature of why this site distinguishes what has been checked from what merely sounds right.

The general base question is harder than it looks. Describing the attractor structure for arbitrary base and digit count is a real research question rather than an exercise, and this page does not attempt it. See the sources below for the base-dependent treatment.