After almost 3 years and no solution for such an easy problem?? note that if in a step, the number of pebbles in each pile is divisible by a positive integer $n$,then in every step,all the numbers are again divisible by $n$.At the begining,all the piles contain odd number of pebbles,so in first step,we should combine 2 of them,resulting one of the pairs $(54,51),(56,49),(5,100)$ which the number of pebbles in each pile will be divisible by 3,7,5,respectively. so at the end,we cannot have 105 piles each with 1 pebble.