Let the original three-digit number be abc=100a+10b+c. According to the first piece of information from the problem, we have
a+b+c=16.
According to the second piece of information, bac=abc−360, thus
100b+10a+c3604=100a+10b+c−360,=90a−90b,=a−b.
According to the third piece of information, acb=abc+54, thus
100a+10c+b9c−9bc−b=100a+10b+c+54,=54,=6.
If we express a=b+4 and c=b+6 from the second and third pieces of information, respectively, and substitute them into the first, we get 3b+10=16, thus b=2. Substituting this result into the previous expressions, we obtain a=6 and c=8. The original three-digit number was 628.
Evaluation. 2 points for each of the equations (2) and (3); 2 points for solving the system and determining the unknown number. A correct solution without further commentary is worth 1 point.
Notes. The differences between numbers formed by swapping two digits are always multiples of nine, with the corresponding multiple reflecting the positions of the swapped digits. For example, in the previous solution, we see abc−bac=90(a−b) and abc−acb=9(b−c), similarly abc−cba=99(a−b). Such or similar reasoning before actually solving the problem allows for a quicker derivation of equations (2) and (3).
The second and third pieces of information from the problem can be visually represented as follows:
abc−bac360acb−abc
By comparing the highest place values, we see that the difference a−b is 3 or 4, and the difference c−b is 5 or 6 (the two possibilities for each difference correspond to whether or not there is a carry-over). These constraints, together with (1), yield a unique solution that can be found by systematically testing the possibilities. For example, the conditions a−b=4 and a+b+c=16 are satisfied by the numbers 952, 844, 736, and 628, of which only the last one also satisfies the constraint on the difference c−b.