Hello Math-Fun, here is a puzzle for the new year -- hope it is sound (the puzzle); 1) Let's say we need 2 "jumps" to rearrange ABCD into CABD, like this: A B C D Jump #1: A C B D Jump #2: C A B D 2) Let's say we always look for the smallest quantity of jumps to rearrange a word "a" into one of its anagrams "b". We'll then need 3 "jumps" to rearrange ABCD into CBAD: A B C D Jump #1: B A C D Jump #2: B C A D Jump #3: C B A D Note: - to transform the word ABCD into CABD you need 2 jumps, BUT to transform the word ABAD into AABD you only need 1 jump. 3) Instead of letter-words "a" and "b", let's consider integers. We will deal only with integers "a" and "b" which obey these simple rules: - a is not equal to b - neither a nor b have a leading zero - both a and b > 9 ----------------------------------------------------------------- Let's start a sequence S of b(n)s where b is the smallest integer which can be produced in n jumps. I guess S starts like this: S = 12, 102, 102, ... b(1) = 12 because "12" is the smallest integer obtainable in 1 jump [the jump 21 -> 12] b(2) = 102 because "102" is the smallest integer obtainable in 2 jumps [210-120-102] b(3) = 102 because "102" is the smallest integer obtainable in 3 jumps [201-210-120-102] etc. Remark 1: we do NOT accept intermediate "integers" with leading zero(es). That is, you cannot produce 102 in 3 jumps like this: 201-021-012-102 Remark 2: we do NOT accept two intermediate "integers" being the same. What is the smallest integer obtainable in 2013 jumps? Best, É.