Base | Representation |
---|---|
bin | 1101000001101110001001… |
… | …1101101100011111111001 |
3 | 1212201021201212010112012221 |
4 | 3100123202131230133321 |
5 | 3334132423410300441 |
6 | 50243555134004041 |
7 | 3005550516413443 |
oct | 320334235543771 |
9 | 55637655115187 |
10 | 14323220400121 |
11 | 4622497568206 |
12 | 1733b29646621 |
13 | 7cb89b6bc589 |
14 | 377367471493 |
15 | 19c8a6e058d1 |
hex | d06e276c7f9 |
14323220400121 has 2 divisors, whose sum is σ = 14323220400122. Its totient is φ = 14323220400120.
The previous prime is 14323220400037. The next prime is 14323220400133. The reversal of 14323220400121 is 12100402232341.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 13826766286096 + 496454114025 = 3718436^2 + 704595^2 .
It is a cyclic number.
It is not a de Polignac number, because 14323220400121 - 211 = 14323220398073 is a prime.
It is not a weakly prime, because it can be changed into another prime (14323220400821) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7161610200060 + 7161610200061.
It is an arithmetic number, because the mean of its divisors is an integer number (7161610200061).
Almost surely, 214323220400121 is an apocalyptic number.
It is an amenable number.
14323220400121 is a deficient number, since it is larger than the sum of its proper divisors (1).
14323220400121 is an equidigital number, since it uses as much as digits as its factorization.
14323220400121 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2304, while the sum is 25.
Adding to 14323220400121 its reverse (12100402232341), we get a palindrome (26423622632462).
The spelling of 14323220400121 in words is "fourteen trillion, three hundred twenty-three billion, two hundred twenty million, four hundred thousand, one hundred twenty-one".
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