Base | Representation |
---|---|
bin | 1011111100010110101101… |
… | …0000000111011000110111 |
3 | 1201111100201211001210121222 |
4 | 2333011223100013120313 |
5 | 3210121320142114201 |
6 | 43532311234500555 |
7 | 2523502111553015 |
oct | 277055320073067 |
9 | 51440654053558 |
10 | 13131514410551 |
11 | 4203053601999 |
12 | 1580b8573475b |
13 | 7433b1529464 |
14 | 3357d66a58b5 |
15 | 17b8aa43e91b |
hex | bf16b407637 |
13131514410551 has 4 divisors (see below), whose sum is σ = 13131705408192. Its totient is φ = 13131323412912.
The previous prime is 13131514410481. The next prime is 13131514410571. The reversal of 13131514410551 is 15501441513131.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13131514410551 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13131514410571) 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, 95395655 + ... + 95533208.
It is an arithmetic number, because the mean of its divisors is an integer number (3282926352048).
Almost surely, 213131514410551 is an apocalyptic number.
13131514410551 is a deficient number, since it is larger than the sum of its proper divisors (190997641).
13131514410551 is an equidigital number, since it uses as much as digits as its factorization.
13131514410551 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 190997640.
The product of its (nonzero) digits is 18000, while the sum is 35.
Adding to 13131514410551 its reverse (15501441513131), we get a palindrome (28632955923682).
The spelling of 13131514410551 in words is "thirteen trillion, one hundred thirty-one billion, five hundred fourteen million, four hundred ten thousand, five hundred fifty-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.066 sec. • engine limits •