Why the Sum Becomes 1089

Write 532 on three separate cards. Swapping the outside cards gives 235. Subtract the smaller number from the larger to obtain 297, reverse that to 792, then add 297 and 792. The sum is 1089. You can pause after each operation and check it without needing a formula.

The rule does not require the middle digit to sit between the other two in size. It requires the outside digits to differ. It also keeps three written positions when the subtraction produces only two digits: 99 must be written as 099 before it is reversed.

Compare 532 and 572. Their reversals are 235 and 275, and both subtractions give 297. Raising the middle digit by four adds forty to each side of the subtraction, leaving the difference unchanged.

The outside digits do matter, but only through their gap. For a gap of three the difference is 99 × 3 = 297. Gaps of one through nine give just nine possible differences. The second view lays these cases out without requiring you to search through every starting number.

Starting with 201 gives 201 − 102 = 99. If you write 099, reversal gives 990 and the final sum is 1089. If you discard the leading zero, reversal gives 99 and the sum is 198. The numerical value of 099 is still ninety-nine, but the digit string is different from 99.

This is also why 1089 should not be described as an attractor here. We performed a fixed list of operations and then stopped. We did not repeatedly feed the sum through the same transformation or prove that it is a fixed point.

Let $a,b,c$ be decimal digits with $a\ne c$, allowing $a=0$ because the laboratory stores three positions. Let $N=100a+10b+c$ and $R=100c+10b+a$. Define $D=|N-R|$, write $D$ in three positions, and let $D^{\mathrm{rev}}$ be the value of that three-position reversal. Then

$D+D^{\mathrm{rev}}=1089.$

When $a=c$, the procedure instead gives zero. That is a case outside the theorem's hypothesis, not a contradiction of its conclusion. All arithmetic here uses base ten.

The middle terms cancel, so $D=99|a-c|$. Put $d=|a-c|$, where $1\le d\le 9$. The identity

$99d=100(d-1)+90+(10-d)$

shows that the three digits of $D$ are $d-1$, $9$ and $10-d$. Their outside sum is nine. Reversing these digits and adding therefore gives

$100((d-1)+(10-d))+180+((d-1)+(10-d))=1089.$

The expression includes $d=1$: its hundreds digit is zero, and retaining that digit is precisely the padding condition. This algebra covers every admissible input, independently of the finite check performed by the tests.

The test suite checks all 1000 three-place strings, including their 100 equal-outside-digit exceptions. That is a complete finite check, while the argument above explains why the check must succeed. Neither calculation establishes a result for longer strings or a different base.

You can investigate those variants by rewriting the place-value argument with another base or another number of positions. For a contrast with a genuinely unresolved reversal process, the 196 exploration repeatedly reverses and adds, and does not inherit the theorem proved here.