Base | Representation |
---|---|
bin | 100000011010011011000111… |
… | …011110001101100111010011 |
3 | 200200201222000100121122000211 |
4 | 200122123013132031213103 |
5 | 122141043140122442212 |
6 | 1223104030450024551 |
7 | 42012064640630524 |
oct | 4032330736154723 |
9 | 620658010548024 |
10 | 142553311140307 |
11 | 41470579866271 |
12 | 13ba3953578157 |
13 | 6170947b45472 |
14 | 272b88d36a84b |
15 | 11732134e59a7 |
hex | 81a6c778d9d3 |
142553311140307 has 2 divisors, whose sum is σ = 142553311140308. Its totient is φ = 142553311140306.
The previous prime is 142553311140293. The next prime is 142553311140311. The reversal of 142553311140307 is 703041113355241.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 142553311140307 - 227 = 142553176922579 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 142553311140307.
It is not a weakly prime, because it can be changed into another prime (142553311140347) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71276655570153 + 71276655570154.
It is an arithmetic number, because the mean of its divisors is an integer number (71276655570154).
Almost surely, 2142553311140307 is an apocalyptic number.
142553311140307 is a deficient number, since it is larger than the sum of its proper divisors (1).
142553311140307 is an equidigital number, since it uses as much as digits as its factorization.
142553311140307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 151200, while the sum is 40.
Adding to 142553311140307 its reverse (703041113355241), we get a palindrome (845594424495548).
The spelling of 142553311140307 in words is "one hundred forty-two trillion, five hundred fifty-three billion, three hundred eleven million, one hundred forty thousand, three hundred seven".
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