Base | Representation |
---|---|
bin | 10101010100011111011… |
… | …111100111100010100111 |
3 | 12012001201012010210110222 |
4 | 111110133133213202213 |
5 | 143001022212330210 |
6 | 3041021355434555 |
7 | 210564563322365 |
oct | 25243737474247 |
9 | 5161635123428 |
10 | 1465112230055 |
11 | 515394727471 |
12 | 1b7b47155a5b |
13 | a820c529056 |
14 | 50caa016635 |
15 | 2819e5d7655 |
hex | 1551f7e78a7 |
1465112230055 has 4 divisors (see below), whose sum is σ = 1758134676072. Its totient is φ = 1172089784040.
The previous prime is 1465112230027. The next prime is 1465112230067. The reversal of 1465112230055 is 5500322115641.
It is a semiprime because it is the product of two primes.
It is not a de Polignac number, because 1465112230055 - 28 = 1465112229799 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 1465112229997 and 1465112230024.
It is a congruent 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, 146511223001 + ... + 146511223010.
It is an arithmetic number, because the mean of its divisors is an integer number (439533669018).
Almost surely, 21465112230055 is an apocalyptic number.
1465112230055 is a deficient number, since it is larger than the sum of its proper divisors (293022446017).
1465112230055 is an equidigital number, since it uses as much as digits as its factorization.
1465112230055 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 293022446016.
The product of its (nonzero) digits is 36000, while the sum is 35.
Adding to 1465112230055 its reverse (5500322115641), we get a palindrome (6965434345696).
The spelling of 1465112230055 in words is "one trillion, four hundred sixty-five billion, one hundred twelve million, two hundred thirty thousand, fifty-five".
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