Base | Representation |
---|---|
bin | 111100100010011000… |
… | …0001110100010110111 |
3 | 110102120002122212111220 |
4 | 1321010300032202313 |
5 | 4112221120341210 |
6 | 135420005001423 |
7 | 12251404056063 |
oct | 1710460164267 |
9 | 412502585456 |
10 | 130002512055 |
11 | 50152035509 |
12 | 212416b3273 |
13 | c34a574060 |
14 | 6413960aa3 |
15 | 35ad1acc70 |
hex | 1e44c0e8b7 |
130002512055 has 16 divisors (see below), whose sum is σ = 224004328800. Its totient is φ = 64001236608.
The previous prime is 130002512053. The next prime is 130002512071. The reversal of 130002512055 is 550215200031.
It is a happy number.
It is not a de Polignac number, because 130002512055 - 21 = 130002512053 is a prime.
It is a super-2 number, since 2×1300025120552 (a number of 23 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (130002512053) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 333339580 + ... + 333339969.
It is an arithmetic number, because the mean of its divisors is an integer number (14000270550).
Almost surely, 2130002512055 is an apocalyptic number.
130002512055 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
130002512055 is a deficient number, since it is larger than the sum of its proper divisors (94001816745).
130002512055 is a wasteful number, since it uses less digits than its factorization.
130002512055 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 666679570.
The product of its (nonzero) digits is 1500, while the sum is 24.
Adding to 130002512055 its reverse (550215200031), we get a palindrome (680217712086).
The spelling of 130002512055 in words is "one hundred thirty billion, two million, five hundred twelve thousand, fifty-five".
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