Base | Representation |
---|---|
bin | 1011010011010101101… |
… | …1011100011011111011 |
3 | 200120012000221200120012 |
4 | 2310311123130123323 |
5 | 11140124444241011 |
6 | 225111143052135 |
7 | 20012466022013 |
oct | 2646533343373 |
9 | 616160850505 |
10 | 194169915131 |
11 | 7538a945145 |
12 | 3176b14104b |
13 | 15405505c0a |
14 | 957daa2c43 |
15 | 50b6701a8b |
hex | 2d356dc6fb |
194169915131 has 2 divisors, whose sum is σ = 194169915132. Its totient is φ = 194169915130.
The previous prime is 194169915109. The next prime is 194169915151. The reversal of 194169915131 is 131519961491.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-194169915131 is a prime.
It is a super-2 number, since 2×1941699151312 (a number of 23 digits) contains 22 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 194169915131.
It is not a weakly prime, because it can be changed into another prime (194169915151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 97084957565 + 97084957566.
It is an arithmetic number, because the mean of its divisors is an integer number (97084957566).
Almost surely, 2194169915131 is an apocalyptic number.
194169915131 is a deficient number, since it is larger than the sum of its proper divisors (1).
194169915131 is an equidigital number, since it uses as much as digits as its factorization.
194169915131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 262440, while the sum is 50.
The spelling of 194169915131 in words is "one hundred ninety-four billion, one hundred sixty-nine million, nine hundred fifteen thousand, one hundred thirty-one".
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