Base | Representation |
---|---|
bin | 11101110010100111110000… |
… | …100010110110000000010111 |
3 | 122011220120201100211112002221 |
4 | 131302213300202312000113 |
5 | 114133131040313433143 |
6 | 1142354403140043211 |
7 | 36412012424136664 |
oct | 3562476042660027 |
9 | 564816640745087 |
10 | 131022013030423 |
11 | 3882513a689749 |
12 | 12840b3b093507 |
13 | 58154272b59c8 |
14 | 244d6dbdb8a6b |
15 | 10232b36642ed |
hex | 7729f08b6017 |
131022013030423 has 2 divisors, whose sum is σ = 131022013030424. Its totient is φ = 131022013030422.
The previous prime is 131022013030417. The next prime is 131022013030471. The reversal of 131022013030423 is 324030310220131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131022013030423 - 237 = 130884574076951 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131022013031423) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65511006515211 + 65511006515212.
It is an arithmetic number, because the mean of its divisors is an integer number (65511006515212).
Almost surely, 2131022013030423 is an apocalyptic number.
131022013030423 is a deficient number, since it is larger than the sum of its proper divisors (1).
131022013030423 is an equidigital number, since it uses as much as digits as its factorization.
131022013030423 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2592, while the sum is 25.
Adding to 131022013030423 its reverse (324030310220131), we get a palindrome (455052323250554).
The spelling of 131022013030423 in words is "one hundred thirty-one trillion, twenty-two billion, thirteen million, thirty thousand, four hundred twenty-three".
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