1 7 2 9 1 7 2 7 1 7 2 5 1 7 2 3 ⋅ ⋅ ⋅ 3 1
Find the last 2 digits of the above number.
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Nice solution. The Carmichael Lamba really does make calculations easier.
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Since λ ( 1 0 0 ) = 2 0 and λ ( 2 0 ) = 4 , where λ denoted the Carmichael's lambda function , the given number is congruent to 2 9 7 1 = ( − 1 + 3 0 ) 7 ≡ − 1 + 7 × 3 0 ≡ 9 ( m o d 1 0 0 )