Base | Representation |
---|---|
bin | 1011111011101100010010… |
… | …1101101001000110100101 |
3 | 1201110021100110002012222011 |
4 | 2332323010231221012211 |
5 | 3204430011341434112 |
6 | 43523145452234221 |
7 | 2522620030340404 |
oct | 276730455510645 |
9 | 51407313065864 |
10 | 13120130421157 |
11 | 41a92416441a1 |
12 | 157a92916b971 |
13 | 7422b8ba7b66 |
14 | 3350367db63b |
15 | 17b440d00aa7 |
hex | beec4b691a5 |
13120130421157 has 2 divisors, whose sum is σ = 13120130421158. Its totient is φ = 13120130421156.
The previous prime is 13120130421149. The next prime is 13120130421227. The reversal of 13120130421157 is 75112403102131.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 8739171651681 + 4380958769476 = 2956209^2 + 2093074^2 .
It is a cyclic number.
It is not a de Polignac number, because 13120130421157 - 23 = 13120130421149 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13120130451157) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6560065210578 + 6560065210579.
It is an arithmetic number, because the mean of its divisors is an integer number (6560065210579).
Almost surely, 213120130421157 is an apocalyptic number.
It is an amenable number.
13120130421157 is a deficient number, since it is larger than the sum of its proper divisors (1).
13120130421157 is an equidigital number, since it uses as much as digits as its factorization.
13120130421157 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5040, while the sum is 31.
Adding to 13120130421157 its reverse (75112403102131), we get a palindrome (88232533523288).
The spelling of 13120130421157 in words is "thirteen trillion, one hundred twenty billion, one hundred thirty million, four hundred twenty-one thousand, one hundred fifty-seven".
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