Base | Representation |
---|---|
bin | 110001000000111001101011… |
… | …1100010100011111100100011 |
3 | 2002112111212011120001201112011 |
4 | 1202001303113202203330203 |
5 | 423002133210413142201 |
6 | 4124535301114413351 |
7 | 156545201416004401 |
oct | 14201632742437443 |
9 | 2075455146051464 |
10 | 431132433334051 |
11 | 115410269021649 |
12 | 4043042b731257 |
13 | 156747c9cb7697 |
14 | 7866cc7762471 |
15 | 34c9b25e40151 |
hex | 1881cd78a3f23 |
431132433334051 has 2 divisors, whose sum is σ = 431132433334052. Its totient is φ = 431132433334050.
The previous prime is 431132433334019. The next prime is 431132433334073. The reversal of 431132433334051 is 150433334231134.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 431132433334051 - 25 = 431132433334019 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 431132433333995 and 431132433334013.
It is not a weakly prime, because it can be changed into another prime (431132433334651) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 215566216667025 + 215566216667026.
It is an arithmetic number, because the mean of its divisors is an integer number (215566216667026).
Almost surely, 2431132433334051 is an apocalyptic number.
431132433334051 is a deficient number, since it is larger than the sum of its proper divisors (1).
431132433334051 is an equidigital number, since it uses as much as digits as its factorization.
431132433334051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 466560, while the sum is 40.
Adding to 431132433334051 its reverse (150433334231134), we get a palindrome (581565767565185).
The spelling of 431132433334051 in words is "four hundred thirty-one trillion, one hundred thirty-two billion, four hundred thirty-three million, three hundred thirty-four thousand, fifty-one".
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