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1 solution
Note that by the Binomial Theorem we have
1 0 6 1 5 2 0 1 5 0 6 0 1 = 1 × 1 0 1 2 + 6 × 1 0 1 0 + 1 5 × 1 0 8 + 2 0 × 1 0 6 + 1 5 × 1 0 4 + 6 × 1 0 2 + 1 × 1 0 0 = ( 0 6 ) 1 0 1 2 + ( 1 6 ) 1 0 1 0 + ( 2 6 ) 1 0 8 + ( 3 6 ) 1 0 6 + ( 4 6 ) 1 0 4 + ( 5 6 ) 1 0 2 + ( 6 6 ) 1 0 0 = ( 1 0 2 + 1 ) 6 = 1 0 1 6 .
Therefore 6 1 0 6 1 5 2 0 1 5 0 6 0 1 = 1 0 1 .