Base | Representation |
---|---|
bin | 11101110101100010010100… |
… | …100000101011001101110011 |
3 | 122012121201202222002221002012 |
4 | 131311202110200223031303 |
5 | 114144421314044432404 |
6 | 1143030412435513135 |
7 | 36432333503643542 |
oct | 3565422440531563 |
9 | 565551688087065 |
10 | 131222332420979 |
11 | 388a2095768788 |
12 | 128739257587ab |
13 | 582b29c831877 |
14 | 24592a267a159 |
15 | 10285d9b77a6e |
hex | 77589482b373 |
131222332420979 has 2 divisors, whose sum is σ = 131222332420980. Its totient is φ = 131222332420978.
The previous prime is 131222332420937. The next prime is 131222332421027. The reversal of 131222332420979 is 979024233222131.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131222332420979 is a prime.
It is a super-3 number, since 3×1312223324209793 (a number of 43 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (131222332425979) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65611166210489 + 65611166210490.
It is an arithmetic number, because the mean of its divisors is an integer number (65611166210490).
Almost surely, 2131222332420979 is an apocalyptic number.
131222332420979 is a deficient number, since it is larger than the sum of its proper divisors (1).
131222332420979 is an equidigital number, since it uses as much as digits as its factorization.
131222332420979 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1959552, while the sum is 50.
The spelling of 131222332420979 in words is "one hundred thirty-one trillion, two hundred twenty-two billion, three hundred thirty-two million, four hundred twenty thousand, nine hundred seventy-nine".
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