Base | Representation |
---|---|
bin | 11011010101001010110110… |
… | …001011001001100111011111 |
3 | 120202121012101211100022101011 |
4 | 123111022312023021213133 |
5 | 111223342204321402033 |
6 | 1103352012552200051 |
7 | 34214210614325056 |
oct | 3325126613114737 |
9 | 522535354308334 |
10 | 120202011122143 |
11 | 35333428689225 |
12 | 11593b56399027 |
13 | 520c0022b8742 |
14 | 2197b462a339d |
15 | dd6ae4437ccd |
hex | 6d52b62c99df |
120202011122143 has 2 divisors, whose sum is σ = 120202011122144. Its totient is φ = 120202011122142.
The previous prime is 120202011121999. The next prime is 120202011122189. The reversal of 120202011122143 is 341221110202021.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 120202011122143 - 29 = 120202011121631 is a prime.
It is a super-3 number, since 3×1202020111221433 (a number of 43 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (120902011122143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60101005561071 + 60101005561072.
It is an arithmetic number, because the mean of its divisors is an integer number (60101005561072).
Almost surely, 2120202011122143 is an apocalyptic number.
120202011122143 is a deficient number, since it is larger than the sum of its proper divisors (1).
120202011122143 is an equidigital number, since it uses as much as digits as its factorization.
120202011122143 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 384, while the sum is 22.
Adding to 120202011122143 its reverse (341221110202021), we get a palindrome (461423121324164).
The spelling of 120202011122143 in words is "one hundred twenty trillion, two hundred two billion, eleven million, one hundred twenty-two thousand, one hundred forty-three".
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