Base | Representation |
---|---|
bin | 1011111100000000000101… |
… | …1111100111010001011111 |
3 | 1201110210001211121221201111 |
4 | 2333000001133213101133 |
5 | 3210021402422333434 |
6 | 43525425104522451 |
7 | 2523165523500442 |
oct | 277000137472137 |
9 | 51423054557644 |
10 | 13125445121119 |
11 | 4200519655a11 |
12 | 157b971020427 |
13 | 742954cb0436 |
14 | 3353bc5bb659 |
15 | 17b6526bb964 |
hex | bf0017e745f |
13125445121119 has 2 divisors, whose sum is σ = 13125445121120. Its totient is φ = 13125445121118.
The previous prime is 13125445121093. The next prime is 13125445121143. The reversal of 13125445121119 is 91112154452131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13125445121119 - 223 = 13125436732511 is a prime.
It is a super-2 number, since 2×131254451211192 (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 (13125445121179) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562722560559 + 6562722560560.
It is an arithmetic number, because the mean of its divisors is an integer number (6562722560560).
Almost surely, 213125445121119 is an apocalyptic number.
13125445121119 is a deficient number, since it is larger than the sum of its proper divisors (1).
13125445121119 is an equidigital number, since it uses as much as digits as its factorization.
13125445121119 is an evil number, because the sum of its binary digits is even.
The product of its digits is 43200, while the sum is 40.
The spelling of 13125445121119 in words is "thirteen trillion, one hundred twenty-five billion, four hundred forty-five million, one hundred twenty-one thousand, one hundred nineteen".
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