Base | Representation |
---|---|
bin | 1100010100010011100111… |
… | …0110101001011111111011 |
3 | 1202221200212010112011021202 |
4 | 3011010321312221133323 |
5 | 3233342032120330410 |
6 | 44445323000524415 |
7 | 2565310140654212 |
oct | 305047166513773 |
9 | 52850763464252 |
10 | 13543002511355 |
11 | 4351611941479 |
12 | 162888418070b |
13 | 77313ac3c085 |
14 | 34b6b04b4079 |
15 | 18744068eca5 |
hex | c5139da97fb |
13543002511355 has 4 divisors (see below), whose sum is σ = 16251603013632. Its totient is φ = 10834402009080.
The previous prime is 13543002511349. The next prime is 13543002511387. The reversal of 13543002511355 is 55311520034531.
It is a semiprime because it is the product of two primes.
It is not a de Polignac number, because 13543002511355 - 230 = 13541928769531 is a prime.
It is a super-3 number, since 3×135430025113553 (a number of 40 digits) contains 333 as substring.
It is a Duffinian number.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1354300251131 + ... + 1354300251140.
It is an arithmetic number, because the mean of its divisors is an integer number (4062900753408).
Almost surely, 213543002511355 is an apocalyptic number.
13543002511355 is a deficient number, since it is larger than the sum of its proper divisors (2708600502277).
13543002511355 is an equidigital number, since it uses as much as digits as its factorization.
13543002511355 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2708600502276.
The product of its (nonzero) digits is 135000, while the sum is 38.
Adding to 13543002511355 its reverse (55311520034531), we get a palindrome (68854522545886).
The spelling of 13543002511355 in words is "thirteen trillion, five hundred forty-three billion, two million, five hundred eleven thousand, three hundred fifty-five".
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