Base | Representation |
---|---|
bin | 100101011001100101… |
… | …1000010100100111011 |
3 | 21200022022122102200022 |
4 | 1022303023002210323 |
5 | 2303441222132011 |
6 | 100521342201055 |
7 | 5542201603142 |
oct | 1126313024473 |
9 | 250268572608 |
10 | 80315427131 |
11 | 31074a68165 |
12 | 1369558318b |
13 | 775c5a47a4 |
14 | 3c5ca06759 |
15 | 21510288db |
hex | 12b32c293b |
80315427131 has 2 divisors, whose sum is σ = 80315427132. Its totient is φ = 80315427130.
The previous prime is 80315427097. The next prime is 80315427151. The reversal of 80315427131 is 13172451308.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-80315427131 is a prime.
It is a super-2 number, since 2×803154271312 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 80315427091 and 80315427100.
It is not a weakly prime, because it can be changed into another prime (80315427151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 40157713565 + 40157713566.
It is an arithmetic number, because the mean of its divisors is an integer number (40157713566).
Almost surely, 280315427131 is an apocalyptic number.
80315427131 is a deficient number, since it is larger than the sum of its proper divisors (1).
80315427131 is an equidigital number, since it uses as much as digits as its factorization.
80315427131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20160, while the sum is 35.
Adding to 80315427131 its reverse (13172451308), we get a palindrome (93487878439).
The spelling of 80315427131 in words is "eighty billion, three hundred fifteen million, four hundred twenty-seven thousand, one hundred thirty-one".
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