Base | Representation |
---|---|
bin | 10111010011101010… |
… | …10110101100010001 |
3 | 1012022001102220012022 |
4 | 23221311112230101 |
5 | 201111312344442 |
6 | 5425353141225 |
7 | 622040564666 |
oct | 135165265421 |
9 | 35261386168 |
10 | 12513012497 |
11 | 53412a6547 |
12 | 25126ba815 |
13 | 124552470b |
14 | 869bdda6d |
15 | 4d38093d2 |
hex | 2e9d56b11 |
12513012497 has 4 divisors (see below), whose sum is σ = 12513270000. Its totient is φ = 12512754996.
The previous prime is 12513012493. The next prime is 12513012509. The reversal of 12513012497 is 79421031521.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 12513012497 - 22 = 12513012493 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (12513012493) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 31247 + ... + 161252.
It is an arithmetic number, because the mean of its divisors is an integer number (3128317500).
Almost surely, 212513012497 is an apocalyptic number.
It is an amenable number.
12513012497 is a deficient number, since it is larger than the sum of its proper divisors (257503).
12513012497 is an equidigital number, since it uses as much as digits as its factorization.
12513012497 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 257502.
The product of its (nonzero) digits is 15120, while the sum is 35.
The spelling of 12513012497 in words is "twelve billion, five hundred thirteen million, twelve thousand, four hundred ninety-seven".
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