Base | Representation |
---|---|
bin | 1011110100110100101111… |
… | …1011111001111101100111 |
3 | 1201000222210111021210112120 |
4 | 2331031023323321331213 |
5 | 3201011340320000403 |
6 | 43353033424121023 |
7 | 2511242102323632 |
oct | 275151373717547 |
9 | 51028714253476 |
10 | 13002140000103 |
11 | 41631a4079521 |
12 | 155ba9a466173 |
13 | 73413413cb77 |
14 | 32d442324c19 |
15 | 1783374a0e53 |
hex | bd34bef9f67 |
13002140000103 has 4 divisors (see below), whose sum is σ = 17336186666808. Its totient is φ = 8668093333400.
The previous prime is 13002140000089. The next prime is 13002140000263. The reversal of 13002140000103 is 30100004120031.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 13002140000103 - 229 = 13001603129191 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13002140000503) 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, 2167023333348 + ... + 2167023333353.
It is an arithmetic number, because the mean of its divisors is an integer number (4334046666702).
Almost surely, 213002140000103 is an apocalyptic number.
13002140000103 is a deficient number, since it is larger than the sum of its proper divisors (4334046666705).
13002140000103 is an equidigital number, since it uses as much as digits as its factorization.
13002140000103 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4334046666704.
The product of its (nonzero) digits is 72, while the sum is 15.
Adding to 13002140000103 its reverse (30100004120031), we get a palindrome (43102144120134).
The spelling of 13002140000103 in words is "thirteen trillion, two billion, one hundred forty million, one hundred three", and thus it is an aban number.
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