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In the first 37 decimal places of (sqrt(3)1)^(sqrt(2)1), there are no primes with more than one digit. Any prime with more than one digit has to end in 1, 3, 7, or 9, and the only 1, 3, 7, or 9 in the first 37 digits of this number is the 7 (two digits after the decimal point). However, 87=3*29, so there are no primes with more than one digit in the first 37 digits of (sqrt(3)1)^(sqrt(2)1). In the first 37 digits, there are 35 consecutive digits without a 1, 3, 7, or 9. The probability of 35 consecutive digits without a 1, 3, 7, or 9 is only 0.000000017. [Jacobs]
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