Base | Representation |
---|---|
bin | 1011000001000001111000… |
… | …0100101100010001100111 |
3 | 1120212220222212112011102202 |
4 | 2300100132010230101213 |
5 | 3041422003423234002 |
6 | 41432153000525115 |
7 | 2360041251001664 |
oct | 260203604542147 |
9 | 46786885464382 |
10 | 12112312321127 |
11 | 394a8915a1007 |
12 | 1437544a8079b |
13 | 69b255c86b35 |
14 | 2dc34bd1aa6b |
15 | 160108020502 |
hex | b041e12c467 |
12112312321127 has 2 divisors, whose sum is σ = 12112312321128. Its totient is φ = 12112312321126.
The previous prime is 12112312321111. The next prime is 12112312321147. The reversal of 12112312321127 is 72112321321121.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12112312321127 - 24 = 12112312321111 is a prime.
It is a super-3 number, since 3×121123123211273 (a number of 40 digits) contains 333 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 12112312321127.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12112312321147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6056156160563 + 6056156160564.
It is an arithmetic number, because the mean of its divisors is an integer number (6056156160564).
Almost surely, 212112312321127 is an apocalyptic number.
12112312321127 is a deficient number, since it is larger than the sum of its proper divisors (1).
12112312321127 is an equidigital number, since it uses as much as digits as its factorization.
12112312321127 is an evil number, because the sum of its binary digits is even.
The product of its digits is 2016, while the sum is 29.
Adding to 12112312321127 its reverse (72112321321121), we get a palindrome (84224633642248).
The spelling of 12112312321127 in words is "twelve trillion, one hundred twelve billion, three hundred twelve million, three hundred twenty-one thousand, one hundred twenty-seven".
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