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Consider the number-table below
2 0 1 6 2 0 1 6 2 0 1 6 2 0 1 6 2 0 1 6 2 0 1 6 … Number of 2 0 1 6 = 2 0 1 8 2 0 1 6 2 0 1 6 … 2 0 1 6
Between these 2 0 1 8 numbers, there will be two, which make the same remainder when it is divided by 2 0 1 7 (because the possible values of the remainder are 0 , 1 , 2 , 3 , 4 , … , 2 0 1 6 ). The differenc between these numbers is: Number of 2 0 1 6 = k 2 0 1 6 2 0 1 6 … 2 0 1 6 − Number of 2 0 1 6 = m 2 0 1 6 2 0 1 6 … 2 0 1 6 = Number of 2 0 1 6 = k − m 2 0 1 6 2 0 1 6 … 2 0 1 6 0 0 … 0 0 = 2 0 1 6 … 2 0 1 6 ∗ 1 0 l Since gcd ( 1 0 l , 2 0 1 7 ) = 1 , 2 0 1 7 ∣ Number of 2 0 1 6 = k − m 2 0 1 6 2 0 1 6 … 2 0 1 6