Base | Representation |
---|---|
bin | 1011000001000110010111… |
… | …0010001010111010011001 |
3 | 1120220001002201101012021111 |
4 | 2300101211302022322121 |
5 | 3041431434422103213 |
6 | 41432512215340321 |
7 | 2360113130136502 |
oct | 260214562127231 |
9 | 46801081335244 |
10 | 12113515425433 |
11 | 3950349728094 |
12 | 143781b9816a1 |
13 | 69b3b82c7286 |
14 | 2dc423a17ba9 |
15 | 160178970a3d |
hex | b0465c8ae99 |
12113515425433 has 2 divisors, whose sum is σ = 12113515425434. Its totient is φ = 12113515425432.
The previous prime is 12113515425373. The next prime is 12113515425437. The reversal of 12113515425433 is 33452451531121.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 8662826519824 + 3450688905609 = 2943268^2 + 1857603^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-12113515425433 is a prime.
It is not a weakly prime, because it can be changed into another prime (12113515425437) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6056757712716 + 6056757712717.
It is an arithmetic number, because the mean of its divisors is an integer number (6056757712717).
Almost surely, 212113515425433 is an apocalyptic number.
It is an amenable number.
12113515425433 is a deficient number, since it is larger than the sum of its proper divisors (1).
12113515425433 is an equidigital number, since it uses as much as digits as its factorization.
12113515425433 is an evil number, because the sum of its binary digits is even.
The product of its digits is 216000, while the sum is 40.
Adding to 12113515425433 its reverse (33452451531121), we get a palindrome (45565966956554).
The spelling of 12113515425433 in words is "twelve trillion, one hundred thirteen billion, five hundred fifteen million, four hundred twenty-five thousand, four hundred thirty-three".
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