Base | Representation |
---|---|
bin | 100011011000011101101000… |
… | …0101100001010011000100111 |
3 | 1111210221212012201210221020011 |
4 | 1012300323100230022120213 |
5 | 311243104130342044201 |
6 | 3021530552415351051 |
7 | 122361203034412432 |
oct | 10660732054123047 |
9 | 1453855181727204 |
10 | 311225421440551 |
11 | 90190a61856132 |
12 | 2aaa56a3b22487 |
13 | 1048757b78bb47 |
14 | 56bd5683bc819 |
15 | 25eaa4a15b151 |
hex | 11b0ed0b0a627 |
311225421440551 has 2 divisors, whose sum is σ = 311225421440552. Its totient is φ = 311225421440550.
The previous prime is 311225421440503. The next prime is 311225421440569. The reversal of 311225421440551 is 155044124522113.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-311225421440551 is a prime.
It is a super-2 number, since 2×3112254214405512 (a number of 30 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (311225421440051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155612710720275 + 155612710720276.
It is an arithmetic number, because the mean of its divisors is an integer number (155612710720276).
Almost surely, 2311225421440551 is an apocalyptic number.
311225421440551 is a deficient number, since it is larger than the sum of its proper divisors (1).
311225421440551 is an equidigital number, since it uses as much as digits as its factorization.
311225421440551 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 192000, while the sum is 40.
Adding to 311225421440551 its reverse (155044124522113), we get a palindrome (466269545962664).
The spelling of 311225421440551 in words is "three hundred eleven trillion, two hundred twenty-five billion, four hundred twenty-one million, four hundred forty thousand, five hundred fifty-one".
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