Base | Representation |
---|---|
bin | 11101100100010101100101… |
… | …000110011111011000010011 |
3 | 122001102202001112021220210012 |
4 | 131210111211012133120103 |
5 | 114021040240240134303 |
6 | 1140323424504425135 |
7 | 36251052600545360 |
oct | 3544254506373023 |
9 | 561382045256705 |
10 | 130040421021203 |
11 | 38486918579780 |
12 | 12702855a781ab |
13 | 57739b3929204 |
14 | 2417dc0643467 |
15 | 10079b2ee75d8 |
hex | 76456519f613 |
130040421021203 has 128 divisors (see below), whose sum is σ = 182196114186240. Its totient is φ = 89696839680000.
The previous prime is 130040421021197. The next prime is 130040421021241. The reversal of 130040421021203 is 302120124040031.
It is not a de Polignac number, because 130040421021203 - 216 = 130040420955667 is a prime.
It is a super-2 number, since 2×1300404210212032 (a number of 29 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (130040421021283) by changing a digit.
It is a polite number, since it can be written in 127 ways as a sum of consecutive naturals, for example, 1016265878 + ... + 1016393828.
It is an arithmetic number, because the mean of its divisors is an integer number (1423407142080).
Almost surely, 2130040421021203 is an apocalyptic number.
130040421021203 is a deficient number, since it is larger than the sum of its proper divisors (52155693165037).
130040421021203 is a wasteful number, since it uses less digits than its factorization.
130040421021203 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 128709.
The product of its (nonzero) digits is 1152, while the sum is 23.
Adding to 130040421021203 its reverse (302120124040031), we get a palindrome (432160545061234).
The spelling of 130040421021203 in words is "one hundred thirty trillion, forty billion, four hundred twenty-one million, twenty-one thousand, two hundred three".
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