Base | Representation |
---|---|
bin | 111100101000100011… |
… | …1111001011000101010 |
3 | 110110002120011120001021 |
4 | 1321101013321120222 |
5 | 4113132232342214 |
6 | 135452341043054 |
7 | 12256504051612 |
oct | 1712107713052 |
9 | 413076146037 |
10 | 130210043434 |
11 | 502491a1949 |
12 | 2129b0b648a |
13 | c3815673a3 |
14 | 6433343a42 |
15 | 35c1503924 |
hex | 1e511f962a |
130210043434 has 4 divisors (see below), whose sum is σ = 195315065154. Its totient is φ = 65105021716.
The previous prime is 130210043399. The next prime is 130210043507. The reversal of 130210043434 is 434340012031.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 116892191025 + 13317852409 = 341895^2 + 115403^2 .
It is a super-2 number, since 2×1302100434342 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 130210043399 and 130210043408.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32552510857 + ... + 32552510860.
Almost surely, 2130210043434 is an apocalyptic number.
130210043434 is a deficient number, since it is larger than the sum of its proper divisors (65105021720).
130210043434 is an equidigital number, since it uses as much as digits as its factorization.
130210043434 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 65105021719.
The product of its (nonzero) digits is 3456, while the sum is 25.
Adding to 130210043434 its reverse (434340012031), we get a palindrome (564550055465).
The spelling of 130210043434 in words is "one hundred thirty billion, two hundred ten million, forty-three thousand, four hundred thirty-four".
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