Base | Representation |
---|---|
bin | 1011000010110100101001… |
… | …1110110011010010000111 |
3 | 1120222212112001121211102211 |
4 | 2300231022132303102013 |
5 | 3042423103231243101 |
6 | 41454250122044251 |
7 | 2362211613044521 |
oct | 260551236632207 |
9 | 46885461554384 |
10 | 12143122134151 |
11 | 39619619685a8 |
12 | 1441503049687 |
13 | 6a1126059528 |
14 | 2dda31b26a11 |
15 | 160d0ccadc51 |
hex | b0b4a7b3487 |
12143122134151 has 2 divisors, whose sum is σ = 12143122134152. Its totient is φ = 12143122134150.
The previous prime is 12143122134083. The next prime is 12143122134167. The reversal of 12143122134151 is 15143122134121.
12143122134151 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 12143122134151 - 27 = 12143122134023 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12143122138151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6071561067075 + 6071561067076.
It is an arithmetic number, because the mean of its divisors is an integer number (6071561067076).
Almost surely, 212143122134151 is an apocalyptic number.
12143122134151 is a deficient number, since it is larger than the sum of its proper divisors (1).
12143122134151 is an equidigital number, since it uses as much as digits as its factorization.
12143122134151 is an evil number, because the sum of its binary digits is even.
The product of its digits is 5760, while the sum is 31.
Adding to 12143122134151 its reverse (15143122134121), we get a palindrome (27286244268272).
The spelling of 12143122134151 in words is "twelve trillion, one hundred forty-three billion, one hundred twenty-two million, one hundred thirty-four thousand, one hundred fifty-one".
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