Base | Representation |
---|---|
bin | 11000100111000000… |
… | …01110001001010001 |
3 | 1021002202221000021111 |
4 | 30103200032021101 |
5 | 204024300144122 |
6 | 10023005253321 |
7 | 645260062165 |
oct | 142340161121 |
9 | 37082830244 |
10 | 13212115537 |
11 | 566a992a55 |
12 | 2688864241 |
13 | 132730b75a |
14 | 8d49c2ca5 |
15 | 524da0d77 |
hex | 31380e251 |
13212115537 has 2 divisors, whose sum is σ = 13212115538. Its totient is φ = 13212115536.
The previous prime is 13212115529. The next prime is 13212115559. The reversal of 13212115537 is 73551121231.
It is a happy number.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 11811559761 + 1400555776 = 108681^2 + 37424^2 .
It is a cyclic number.
It is not a de Polignac number, because 13212115537 - 23 = 13212115529 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13212115499 and 13212115508.
It is not a weakly prime, because it can be changed into another prime (13212115337) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6606057768 + 6606057769.
It is an arithmetic number, because the mean of its divisors is an integer number (6606057769).
Almost surely, 213212115537 is an apocalyptic number.
It is an amenable number.
13212115537 is a deficient number, since it is larger than the sum of its proper divisors (1).
13212115537 is an equidigital number, since it uses as much as digits as its factorization.
13212115537 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 6300, while the sum is 31.
Adding to 13212115537 its reverse (73551121231), we get a palindrome (86763236768).
The spelling of 13212115537 in words is "thirteen billion, two hundred twelve million, one hundred fifteen thousand, five hundred thirty-seven".
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