Maths Olympiad Prep

Library / Estonian Math Competitions /104 of 148

Estonian Math Competitions

Algebra Difficulty 6.3 National Olympiad Prove it Estonia

(a) Find the largest number expressible as the difference of two two-digit numbers obtained from each other by changing the order of digits.
(b) The same question with three-digit instead of two-digit numbers.

Solution

(a) Let the given two-digit number be ab‾\overline{ab}. The only number that can be obtained by changing the order of digits is ba‾\overline{ba}. The difference of these numbers is (10a+b)−(10b+a)=9(a−b)(10a + b) - (10b + a) = 9(a - b). To obtain the largest difference, aa must be as large as possible and bb as small as possible. Since b=0b = 0 is impossible, we must have a=9a = 9 and b=1b = 1 giving a−b=8a - b = 8. Hence the desired largest difference is 7272.

(b) Let the given three-digit number be abc‾\overline{abc}. Interchanging the last two digits can change it by less than 100100. Interchanging the first two digits can change the number by at most 720720 by part (a) of the problem. It remains to study cases where changing the order of digits results in cab‾\overline{cab}, bca‾\overline{bca} or cba‾\overline{cba}.

* The difference of numbers abc‾\overline{abc} and cab‾\overline{cab} is (100a+10b+c)−(100c+10a+b)=9(10a+b−11c)(100a + 10b + c) - (100c + 10a + b) = 9(10a + b - 11c). To obtain the largest difference, aa and bb must be as large as possible and cc as small as possible. Since cc is the first digit of the number, c=0c = 0 is impossible, whence the largest difference is obtained if a=b=9a = b = 9 and c=1c = 1. This difference is 991−199=792991 - 199 = 792.

* The difference of numbers abc‾\overline{abc} and bca‾\overline{bca} is (100a+10b+c)−(100b+10c+a)=9(11a−10b−c)(100a + 10b + c) - (100b + 10c + a) = 9(11a - 10b - c). To obtain the largest difference, aa must be as large as possible and both bb and cc as small as possible, i.e., a=9a = 9, b=1b = 1 and c=0c = 0. This difference is 910−109=801910 - 109 = 801.

* The difference of numbers abc‾\overline{abc} and cba‾\overline{cba} is (100a+10b+c)−(100c+10b+a)=99(a−c)(100a + 10b + c) - (100c + 10b + a) = 99(a - c). To obtain the largest difference, aa must be as large as possible and cc as small as possible. Since c≠0c \neq 0, the largest difference is obtained if a=9a = 9 and c=1c = 1. Then the difference of the three-digit numbers is 99⋅8=79299 \cdot 8 = 792.

Consequently, the desired largest difference is 801801.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.