Base | Representation |
---|---|
bin | 11101110100011110000011… |
… | …000100011000100110101001 |
3 | 122012100201120001220202221112 |
4 | 131310132003010120212221 |
5 | 114142221200442114014 |
6 | 1142533010330520105 |
7 | 36424130052052664 |
oct | 3564360304304651 |
9 | 565321501822845 |
10 | 131149025348009 |
11 | 38873a97840859 |
12 | 12861687296035 |
13 | 58243b91c9706 |
14 | 245590a7666db |
15 | 102674906873e |
hex | 7747831189a9 |
131149025348009 has 2 divisors, whose sum is σ = 131149025348010. Its totient is φ = 131149025348008.
The previous prime is 131149025347973. The next prime is 131149025348023. The reversal of 131149025348009 is 900843520941131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 130899844675609 + 249180672400 = 11441147^2 + 499180^2 .
It is a cyclic number.
It is not a de Polignac number, because 131149025348009 - 212 = 131149025343913 is a prime.
It is not a weakly prime, because it can be changed into another prime (131149025348069) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65574512674004 + 65574512674005.
It is an arithmetic number, because the mean of its divisors is an integer number (65574512674005).
Almost surely, 2131149025348009 is an apocalyptic number.
It is an amenable number.
131149025348009 is a deficient number, since it is larger than the sum of its proper divisors (1).
131149025348009 is an equidigital number, since it uses as much as digits as its factorization.
131149025348009 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 933120, while the sum is 50.
The spelling of 131149025348009 in words is "one hundred thirty-one trillion, one hundred forty-nine billion, twenty-five million, three hundred forty-eight thousand, nine".
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