Base | Representation |
---|---|
bin | 11101110100000011111100… |
… | …110011100111000010111001 |
3 | 122012021001020010221000002012 |
4 | 131310003330303213002321 |
5 | 114141241303231444433 |
6 | 1142512053542415305 |
7 | 36422115455040236 |
oct | 3564037463470271 |
9 | 565231203830065 |
10 | 131121002999993 |
11 | 3886311796aa56 |
12 | 12858166731535 |
13 | 58218717858b6 |
14 | 245440cc3a78d |
15 | 1025b58d8a848 |
hex | 7740fcce70b9 |
131121002999993 has 2 divisors, whose sum is σ = 131121002999994. Its totient is φ = 131121002999992.
The previous prime is 131121002999989. The next prime is 131121003000059. The reversal of 131121002999993 is 399999200121131.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 127174157231104 + 3946845768889 = 11277152^2 + 1986667^2 .
It is a cyclic number.
It is not a de Polignac number, because 131121002999993 - 22 = 131121002999989 is a prime.
It is not a weakly prime, because it can be changed into another prime (131121002929993) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65560501499996 + 65560501499997.
It is an arithmetic number, because the mean of its divisors is an integer number (65560501499997).
Almost surely, 2131121002999993 is an apocalyptic number.
It is an amenable number.
131121002999993 is a deficient number, since it is larger than the sum of its proper divisors (1).
131121002999993 is an equidigital number, since it uses as much as digits as its factorization.
131121002999993 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2125764, while the sum is 59.
The spelling of 131121002999993 in words is "one hundred thirty-one trillion, one hundred twenty-one billion, two million, nine hundred ninety-nine thousand, nine hundred ninety-three".
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