Base | Representation |
---|---|
bin | 10111010000000101110011… |
… | …00101111100111011000011 |
3 | 20201001000111210111212121122 |
4 | 23220002321211330323003 |
5 | 23200210032013103121 |
6 | 300424552235142455 |
7 | 13525023041460263 |
oct | 1350027145747303 |
9 | 221030453455548 |
10 | 51130404425411 |
11 | 15323312370993 |
12 | 58994b8ba4a2b |
13 | 226b7655c3ccc |
14 | c8aa20b137a3 |
15 | 5da0449c45ab |
hex | 2e80b997cec3 |
51130404425411 has 2 divisors, whose sum is σ = 51130404425412. Its totient is φ = 51130404425410.
The previous prime is 51130404425383. The next prime is 51130404425437. The reversal of 51130404425411 is 11452440403115.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 51130404425411 - 234 = 51113224556227 is a prime.
It is a super-2 number, since 2×511304044254112 (a number of 28 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (51130404425461) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25565202212705 + 25565202212706.
It is an arithmetic number, because the mean of its divisors is an integer number (25565202212706).
Almost surely, 251130404425411 is an apocalyptic number.
51130404425411 is a deficient number, since it is larger than the sum of its proper divisors (1).
51130404425411 is an equidigital number, since it uses as much as digits as its factorization.
51130404425411 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 38400, while the sum is 35.
Adding to 51130404425411 its reverse (11452440403115), we get a palindrome (62582844828526).
The spelling of 51130404425411 in words is "fifty-one trillion, one hundred thirty billion, four hundred four million, four hundred twenty-five thousand, four hundred eleven".
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