Base | Representation |
---|---|
bin | 1100001111010001000100… |
… | …1110011110000001011111 |
3 | 1202122102101011011022021021 |
4 | 3003310101032132001133 |
5 | 3230432222403402111 |
6 | 44341443354145011 |
7 | 2556123445615024 |
oct | 303642116360137 |
9 | 52572334138237 |
10 | 13456421544031 |
11 | 4318922176224 |
12 | 1613b40422167 |
13 | 767c2057796b |
14 | 3474196d954b |
15 | 185074579d71 |
hex | c3d1139e05f |
13456421544031 has 2 divisors, whose sum is σ = 13456421544032. Its totient is φ = 13456421544030.
The previous prime is 13456421544019. The next prime is 13456421544053. The reversal of 13456421544031 is 13044512465431.
13456421544031 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13456421544031 is a prime.
It is a super-2 number, since 2×134564215440312 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13456421544011) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6728210772015 + 6728210772016.
It is an arithmetic number, because the mean of its divisors is an integer number (6728210772016).
Almost surely, 213456421544031 is an apocalyptic number.
13456421544031 is a deficient number, since it is larger than the sum of its proper divisors (1).
13456421544031 is an equidigital number, since it uses as much as digits as its factorization.
13456421544031 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 691200, while the sum is 43.
The spelling of 13456421544031 in words is "thirteen trillion, four hundred fifty-six billion, four hundred twenty-one million, five hundred forty-four thousand, thirty-one".
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