Find the sum of the last two digits of
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I assume the number above 1 9 is 1 7 as you have listed the first 1 0 prime numbers in ascending order.
The trick is to apply Euler's Totient Function repeatedly
φ ( 1 0 0 ) = 4 0 , φ ( 4 0 ) = 1 6 , φ ( 1 0 0 ) = 8
And gcd ( 2 9 , 1 0 0 ) = gcd ( 2 3 , 4 0 ) = gcd ( 1 9 , 1 6 ) = gcd ( 1 7 , 8 ) = 1
2 9 2 3 1 9 1 7 . . 7 5 3 2 ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 1 0 0 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 4 0 ) ( m o d 1 0 0 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 1 6 ) ( m o d 4 0 ) ( m o d 1 0 0 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 8 ) ( m o d 1 6 ) ( m o d 4 0 ) ( m o d 1 0 0 ) 2 9 2 3 1 9 1 ( m o d 1 6 ) ( m o d 4 0 ) ( m o d 1 0 0 ) 2 9 2 3 3 ( m o d 4 0 ) ( m o d 1 0 0 ) 2 9 7 ( m o d 1 0 0 ) 9