Base | Representation |
---|---|
bin | 10111001111111011000011… |
… | …01000001010010010000011 |
3 | 20201000110121121202201010222 |
4 | 23213331201220022102003 |
5 | 23200111222203021430 |
6 | 300422155451213255 |
7 | 13524430036563215 |
oct | 1347754150122203 |
9 | 221013547681128 |
10 | 51124633642115 |
11 | 15320920968aa0 |
12 | 589836844722b |
13 | 226b055b71b37 |
14 | c8a6345305b5 |
15 | 5d9d080679e5 |
hex | 2e7f61a0a483 |
51124633642115 has 16 divisors (see below), whose sum is σ = 67869424025856. Its totient is φ = 36657867909600.
The previous prime is 51124633642099. The next prime is 51124633642127.
51124633642115 is nontrivially palindromic in base 10.
It is not a de Polignac number, because 51124633642115 - 24 = 51124633642099 is a prime.
It is a super-3 number, since 3×511246336421153 (a number of 42 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a Duffinian number.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 6546043937 + ... + 6546051746.
It is an arithmetic number, because the mean of its divisors is an integer number (4241839001616).
Almost surely, 251124633642115 is an apocalyptic number.
51124633642115 is a gapful number since it is divisible by the number (55) formed by its first and last digit.
51124633642115 is a deficient number, since it is larger than the sum of its proper divisors (16744790383741).
51124633642115 is a wasteful number, since it uses less digits than its factorization.
51124633642115 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 13092095770.
The product of its digits is 518400, while the sum is 44.
It can be divided in two parts, 5112463 and 3642115, that added together give a palindrome (8754578).
The spelling of 51124633642115 in words is "fifty-one trillion, one hundred twenty-four billion, six hundred thirty-three million, six hundred forty-two thousand, one hundred fifteen".
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