Base | Representation |
---|---|
bin | 111011110110101000100… |
… | …010000110000000000001 |
3 | 112120012122210101121212112 |
4 | 323312220202012000001 |
5 | 1014342122432241113 |
6 | 12425312011505105 |
7 | 603106505306435 |
oct | 73665042060001 |
9 | 15505583347775 |
10 | 4113111212033 |
11 | 13463a706139a |
12 | 565194758195 |
13 | 23ab31a47ab5 |
14 | 10310b7d89c5 |
15 | 71ed12b35a8 |
hex | 3bda8886001 |
4113111212033 has 2 divisors, whose sum is σ = 4113111212034. Its totient is φ = 4113111212032.
The previous prime is 4113111211969. The next prime is 4113111212047. The reversal of 4113111212033 is 3302121113114.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2771748830449 + 1341362381584 = 1664857^2 + 1158172^2 .
It is a cyclic number.
It is not a de Polignac number, because 4113111212033 - 26 = 4113111211969 is a prime.
It is not a weakly prime, because it can be changed into another prime (4113111212053) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2056555606016 + 2056555606017.
It is an arithmetic number, because the mean of its divisors is an integer number (2056555606017).
Almost surely, 24113111212033 is an apocalyptic number.
It is an amenable number.
4113111212033 is a deficient number, since it is larger than the sum of its proper divisors (1).
4113111212033 is an equidigital number, since it uses as much as digits as its factorization.
4113111212033 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 432, while the sum is 23.
Adding to 4113111212033 its reverse (3302121113114), we get a palindrome (7415232325147).
The spelling of 4113111212033 in words is "four trillion, one hundred thirteen billion, one hundred eleven million, two hundred twelve thousand, thirty-three".
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