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In base 10, there is thought to be no number with a multiplicative persistence greater than 11; this is known to be true for numbers up to 2.67×10 30000. [1] [2] The smallest numbers with persistence 0, 1, 2, ... are: 0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899. (sequence A003001 in the OEIS)
If the value after the double arrow is a very large number itself, the above can recursively be applied to that value. Examples: 10 ↑ ↑ 10 10 10 3.81 × 10 17 {\displaystyle 10\uparrow \uparrow 10^{\,\!10^{10^{3.81\times 10^{17}}}}} (between 10 ↑ ↑ ↑ 2 {\displaystyle 10\uparrow \uparrow \uparrow 2} and 10 ↑ ↑ ↑ 3 {\displaystyle ...
E.g. if a slow processor can solve problems of size x in time t, then a processor twice as fast could only solve problems of size x + constant in the same time t. So exponentially complex algorithms are most often impractical, and the search for more efficient algorithms is one of the central goals of computer science today.
Different values of k give different values of unless w is a rational number, that is, there is an integer d such that dw is an integer. This results from the periodicity of the exponential function, more specifically, that e a = e b {\displaystyle e^{a}=e^{b}} if and only if a − b {\displaystyle a-b} is an integer multiple of 2 π i ...
The fundamental theorem can be derived from Book VII, propositions 30, 31 and 32, and Book IX, proposition 14 of Euclid's Elements.. If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers.
It is possible to extend the problem to ask how many people in a group are necessary for there to be a greater than 50% probability that at least 3, 4, 5, etc. of the group share the same birthday. The first few values are as follows: >50% probability of 3 people sharing a birthday - 88 people; >50% probability of 4 people sharing a birthday ...
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This behavior can be switched of by setting the formula in parentheses: = ( 1 + 2^-52 - 1 ). You will see that even that small value survives. Smaller values will pass away as there are only 53 bits to represent the value, for this case 1.0000000000 0000000000 0000000000 0000000000 0000000000 01, the first representing the 1, and the last the 2 ...