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1 solution
N = 1 1 1 0 − 1 = ( 1 0 + 1 ) 1 0 − 1 = 1 0 1 0 + 1 0 ( 1 0 9 ) + . . . + 1 0 ( 1 0 ) + 1 − 1
Each of the terms 1 0 1 0 + 1 0 ( 1 0 9 ) + . . . + 1 0 ( 1 0 ) ends in at least two zeros. Consequently, there is a common factor of 1 0 0 so that N ends in 0 0 .