What are the last 7 digits of 3 1 1 , 0 0 0 , 0 0 1 ?
Note: Ignore initial zeroes of the answer for example if the answer is 0 0 0 2 4 5 6 , submit 2 4 5 6 .
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The Carmichael's lambda function of 1 0 7 is 5 × 1 0 5 .
Thus, 3 5 × 1 0 5 ≡ 1 ( m o d 1 0 7 ) . Raise both sides to the power of 22 yields 3 1 1 × 1 0 6 ≡ 1 ( m o d 1 0 7 ) . Now multiply both sides by 3, and the answer follows.
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3 1 1 , 0 0 0 , 0 0 1 ≡ 3 × 3 2 × 5 , 5 0 0 , 0 0 0 (mod 1 0 7 ) ≡ 3 × 9 5 , 5 0 0 , 0 0 0 (mod 1 0 7 ) ≡ 3 ( 1 0 − 1 ) 5 , 5 0 0 , 0 0 0 (mod 1 0 7 ) ≡ 3 ( ⋯ + 5 , 5 0 0 , 0 0 0 ( 5 , 4 9 9 , 9 9 9 ) ( 5 0 ) − 5 5 , 0 0 0 , 0 0 0 + 1 ) (mod 1 0 7 ) ≡ 3 ( ⋯ + 5 5 , 0 0 0 , 0 0 0 ( 2 7 , 4 9 9 , 9 9 5 − 1 ) + 1 ) (mod 1 0 7 ) ≡ 3 ( ⋯ + 5 5 , 0 0 0 , 0 0 0 ( 2 7 , 4 9 9 , 9 9 4 ) + 1 ) (mod 1 0 7 ) ≡ 3 ( ⋯ + 1 1 0 , 0 0 0 , 0 0 0 ( 1 3 , 7 4 9 , 9 9 7 ) + 1 ) (mod 1 0 7 ) ≡ 3 (mod 1 0 7 )