Base | Representation |
---|---|
bin | 1010111010100100101100… |
… | …1100111101011001011001 |
3 | 1120111022201020000101200221 |
4 | 2232221023030331121121 |
5 | 3033112331440012213 |
6 | 41305211151400041 |
7 | 2346033613331656 |
oct | 256511314753131 |
9 | 46438636011627 |
10 | 12001400313433 |
11 | 3907846668033 |
12 | 1419b50823021 |
13 | 69095b84ac0b |
14 | 2d6c29972c2d |
15 | 15c2b5d2838d |
hex | aea4b33d659 |
12001400313433 has 16 divisors (see below), whose sum is σ = 12263179057920. Its totient is φ = 11739891454336.
The previous prime is 12001400313403. The next prime is 12001400313443. The reversal of 12001400313433 is 33431300410021.
It is a cyclic number.
It is not a de Polignac number, because 12001400313433 - 213 = 12001400305241 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 12001400313398 and 12001400313407.
It is not an unprimeable number, because it can be changed into a prime (12001400313403) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 978182823 + ... + 978195091.
It is an arithmetic number, because the mean of its divisors is an integer number (766448691120).
Almost surely, 212001400313433 is an apocalyptic number.
It is an amenable number.
12001400313433 is a deficient number, since it is larger than the sum of its proper divisors (261778744487).
12001400313433 is a wasteful number, since it uses less digits than its factorization.
12001400313433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 21536.
The product of its (nonzero) digits is 2592, while the sum is 25.
Adding to 12001400313433 its reverse (33431300410021), we get a palindrome (45432700723454).
The spelling of 12001400313433 in words is "twelve trillion, one billion, four hundred million, three hundred thirteen thousand, four hundred thirty-three".
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