cs=.29 cs | 15 ^ 22 5 3 20 15 18 0 10 11 0 0 0 The zeros in the above indicate that there is a problem, namely that numbers such as 15^22 are very large and are approximated when represented as floating-point values. The residue of the approximate value is not useful. This is easily solved since
Henry Rich On 7/29/2018 2:31 PM, Brian Schott wrote:
cs=.29 15(cs&|@^)22 5 3 20 15 18 0 10 11 0 0 0 The zeros in the above indicate that there is a problem, namely that numbers such as 1522 are very large and exceed the capacity of the computer. This is easily solved since
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