Originally posted by David113
[fen]8/2PPPPPP/3PPPK1/4PP1P/4P1K1/5P1P/8/8 w - - 0 1[/fen]
Paint some pieces black to make the position legal.
bKg6 implies wKg4, but then Pf5/h5 can't be the same color (one K is in check from 2P at once) or different colors (both K's in check). wK must be on g6.
The Black a-pawn can't possibly be on the board. Of the 15 pawns, 7 are Black and 8 White. White is allowed 8 P captures and Black 7.
White needs a minimum of 6 P captures just to restore the a-d file pawns, and that assumes Pc7d7d6 and at least one e-pawn are White. Pf7 and h7 must be White to avoid imaginary check.
The logic of the solution is not too hard to see. Pe7 can't be Black, for this shuts in Bf8 and deprives White of a vital P capture. wPf6/bPf5 is impossible, for one of the White pawns must stay put on the f-file, so they cannot both be stuck behind a bP. bPh3/wPh5 fails because the h-pawns are stuck behind each other, demanding further captures.