Base | Representation |
---|---|
bin | 101000001000000101001… |
… | …1011111101000110010101 |
3 | 201112012221202222020102212 |
4 | 1100100022123331012111 |
5 | 1210324020223431443 |
6 | 15421303251243205 |
7 | 1106303454652544 |
oct | 120201233750625 |
9 | 21465852866385 |
10 | 5514913108373 |
11 | 1836952749009 |
12 | 7509b12a9505 |
13 | 310091a56c06 |
14 | 150ccd45215b |
15 | 986c7703618 |
hex | 5040a6fd195 |
5514913108373 has 2 divisors, whose sum is σ = 5514913108374. Its totient is φ = 5514913108372.
The previous prime is 5514913108351. The next prime is 5514913108379. The reversal of 5514913108373 is 3738013194155.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4085661775204 + 1429251333169 = 2021302^2 + 1195513^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5514913108373 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5514913108379) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2757456554186 + 2757456554187.
It is an arithmetic number, because the mean of its divisors is an integer number (2757456554187).
Almost surely, 25514913108373 is an apocalyptic number.
It is an amenable number.
5514913108373 is a deficient number, since it is larger than the sum of its proper divisors (1).
5514913108373 is an equidigital number, since it uses as much as digits as its factorization.
5514913108373 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1360800, while the sum is 50.
The spelling of 5514913108373 in words is "five trillion, five hundred fourteen billion, nine hundred thirteen million, one hundred eight thousand, three hundred seventy-three".
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