Hello Math-Fun, We are looking for an 18-digit integer (like 946131483695200285) where we have "d" digits between two d's (here: one digit between two 1's, zero digit between the two 0's, nine digits between the two 9's, etc.) Those 18 digits form thus 9 pairs of digits -- the 9 pairs being different one from another. Now cut this integer in chunks so to make a finite monotonically increasing sequence (like this one, for instance, among others): 9,46,131,483,695,200285 We sum the terms: 9+46+131+483+695+200285 = 201649 Question: Find the integer which, properly chunked, will give the smallest possible sum. --- Example #2: 849362432856900151 could give: 8+49+362+432+856+900151 = 901858 ... but if we move the first digit "8" to the end (which is still sound), 849362432856900151 becomes 493624328569001518 ... which, properly chunked, will give: 4+9+362+432+856+900+1518 = 4080 This is my personal best so far. Regards, E. --- P.S. a 20-digit integer such as described is impossible for parity reasons. I'll be delighted to receive the full b-file list of all such 18-digit integers (if someone would be kind enough to compute it).