Base | Representation |
---|---|
bin | 10011100110011011110000… |
… | …00111110101011111111111 |
3 | 12122121111211100000200220111 |
4 | 21303031320013311133333 |
5 | 21122140410101313143 |
6 | 231400442505530451 |
7 | 12036006020123263 |
oct | 1163157007653777 |
9 | 178544740020814 |
10 | 43102012135423 |
11 | 12808508201a8a |
12 | 4a015607a3a27 |
13 | 1b0866b73cc77 |
14 | a9021176a0a3 |
15 | 4eb2b015899d |
hex | 2733781f57ff |
43102012135423 has 4 divisors (see below), whose sum is σ = 43133404716648. Its totient is φ = 43070619554200.
The previous prime is 43102012135393. The next prime is 43102012135439. The reversal of 43102012135423 is 32453121020134.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 43102012135423 - 213 = 43102012127231 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (43102012135123) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 15696288553 + ... + 15696291298.
It is an arithmetic number, because the mean of its divisors is an integer number (10783351179162).
Almost surely, 243102012135423 is an apocalyptic number.
43102012135423 is a deficient number, since it is larger than the sum of its proper divisors (31392581225).
43102012135423 is a wasteful number, since it uses less digits than its factorization.
43102012135423 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 31392581224.
The product of its (nonzero) digits is 17280, while the sum is 31.
Adding to 43102012135423 its reverse (32453121020134), we get a palindrome (75555133155557).
The spelling of 43102012135423 in words is "forty-three trillion, one hundred two billion, twelve million, one hundred thirty-five thousand, four hundred twenty-three".
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