Base | Representation |
---|---|
bin | 100110101001010101010… |
… | …0000110111101001100111 |
3 | 200210202202102000200102121 |
4 | 1031102222200313221213 |
5 | 1144010312210234002 |
6 | 15144013023250411 |
7 | 1055511304545433 |
oct | 115225240675147 |
9 | 20722672020377 |
10 | 5311440321127 |
11 | 1768629657406 |
12 | 719486749a07 |
13 | 2c6b36108855 |
14 | 14510a0579c3 |
15 | 9326944a237 |
hex | 4d4aa837a67 |
5311440321127 has 2 divisors, whose sum is σ = 5311440321128. Its totient is φ = 5311440321126.
The previous prime is 5311440321113. The next prime is 5311440321281. The reversal of 5311440321127 is 7211230441135.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5311440321127 - 27 = 5311440320999 is a prime.
It is a super-2 number, since 2×53114403211272 (a number of 26 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 5311440321092 and 5311440321101.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5311440321727) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2655720160563 + 2655720160564.
It is an arithmetic number, because the mean of its divisors is an integer number (2655720160564).
Almost surely, 25311440321127 is an apocalyptic number.
5311440321127 is a deficient number, since it is larger than the sum of its proper divisors (1).
5311440321127 is an equidigital number, since it uses as much as digits as its factorization.
5311440321127 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 34.
The spelling of 5311440321127 in words is "five trillion, three hundred eleven billion, four hundred forty million, three hundred twenty-one thousand, one hundred twenty-seven".
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