Base | Representation |
---|---|
bin | 11101110011100110111011… |
… | …111000011101001010100011 |
3 | 122012011001210222010211001201 |
4 | 131303212323320131022203 |
5 | 114140234002314331011 |
6 | 1142445502323542031 |
7 | 36416633443166056 |
oct | 3563467370351243 |
9 | 565131728124051 |
10 | 131089848980131 |
11 | 388509902757a8 |
12 | 1285211122a917 |
13 | 581b9482a7b12 |
14 | 2452ad5430a9d |
15 | 1024e33ca9cc1 |
hex | 7739bbe1d2a3 |
131089848980131 has 4 divisors (see below), whose sum is σ = 131092006649872. Its totient is φ = 131087691310392.
The previous prime is 131089848980059. The next prime is 131089848980137.
131089848980131 is nontrivially palindromic in base 10.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 131089848980131 - 217 = 131089848849059 is a prime.
It is a super-2 number, since 2×1310898489801312 (a number of 29 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (131089848980137) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1078743735 + ... + 1078865248.
It is an arithmetic number, because the mean of its divisors is an integer number (32773001662468).
Almost surely, 2131089848980131 is an apocalyptic number.
131089848980131 is a deficient number, since it is larger than the sum of its proper divisors (2157669741).
131089848980131 is an equidigital number, since it uses as much as digits as its factorization.
131089848980131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2157669740.
The product of its (nonzero) digits is 11943936, while the sum is 64.
The spelling of 131089848980131 in words is "one hundred thirty-one trillion, eighty-nine billion, eight hundred forty-eight million, nine hundred eighty thousand, one hundred thirty-one".
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