Base | Representation |
---|---|
bin | 1010111100110101011001… |
… | …0011001110111001101011 |
3 | 1120122000222010000112101001 |
4 | 2233031112103032321223 |
5 | 3034231403222304120 |
6 | 41335113242535431 |
7 | 2351610254460220 |
oct | 257152623167153 |
9 | 46560863015331 |
10 | 12040241213035 |
11 | 3922268304862 |
12 | 1425590507577 |
13 | 69450b729149 |
14 | 2d8a72219947 |
15 | 15d2daba2e0a |
hex | af3564cee6b |
12040241213035 has 64 divisors (see below), whose sum is σ = 16921959505920. Its totient is φ = 8054397388800.
The previous prime is 12040241213029. The next prime is 12040241213087. The reversal of 12040241213035 is 53031214204021.
It is not a de Polignac number, because 12040241213035 - 29 = 12040241212523 is a prime.
It is a super-2 number, since 2×120402412130352 (a number of 27 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 12040241212994 and 12040241213012.
It is an unprimeable number.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 88235110 + ... + 88371460.
It is an arithmetic number, because the mean of its divisors is an integer number (264405617280).
Almost surely, 212040241213035 is an apocalyptic number.
12040241213035 is a deficient number, since it is larger than the sum of its proper divisors (4881718292885).
12040241213035 is a wasteful number, since it uses less digits than its factorization.
12040241213035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 136828.
The product of its (nonzero) digits is 5760, while the sum is 28.
Adding to 12040241213035 its reverse (53031214204021), we get a palindrome (65071455417056).
The spelling of 12040241213035 in words is "twelve trillion, forty billion, two hundred forty-one million, two hundred thirteen thousand, thirty-five".
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