Base | Representation |
---|---|
bin | 10011101100100110110111… |
… | …11100001110101101000011 |
3 | 12200100210100220102111211120 |
4 | 21312103123330032231003 |
5 | 21134124324343241120 |
6 | 232042124042443323 |
7 | 12060225526044240 |
oct | 1166233374165503 |
9 | 180323326374746 |
10 | 43314140212035 |
11 | 1288a472a526a0 |
12 | 4a366a1b03543 |
13 | 1b226765b7b24 |
14 | a9a5b64885c7 |
15 | 501a781ec140 |
hex | 2764dbf0eb43 |
43314140212035 has 32 divisors (see below), whose sum is σ = 86403271905792. Its totient is φ = 18000681646080.
The previous prime is 43314140212009. The next prime is 43314140212067. The reversal of 43314140212035 is 53021204141334.
It is not a de Polignac number, because 43314140212035 - 222 = 43314136017731 is a prime.
It is a super-2 number, since 2×433141402120352 (a number of 28 digits) contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (33).
It is a junction number, because it is equal to n+sod(n) for n = 43314140211987 and 43314140212005.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 18750708894 + ... + 18750711203.
It is an arithmetic number, because the mean of its divisors is an integer number (2700102247056).
Almost surely, 243314140212035 is an apocalyptic number.
43314140212035 is a deficient number, since it is larger than the sum of its proper divisors (43089131693757).
43314140212035 is a wasteful number, since it uses less digits than its factorization.
43314140212035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 37501420123.
The product of its (nonzero) digits is 34560, while the sum is 33.
Adding to 43314140212035 its reverse (53021204141334), we get a palindrome (96335344353369).
The spelling of 43314140212035 in words is "forty-three trillion, three hundred fourteen billion, one hundred forty million, two hundred twelve thousand, thirty-five".
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