Base | Representation |
---|---|
bin | 1011111100110111011111… |
… | …0000011011101100101011 |
3 | 1201112012110002022112221021 |
4 | 2333031313300123230223 |
5 | 3210242330410043111 |
6 | 43540324343353311 |
7 | 2524233143215141 |
oct | 277156760335453 |
9 | 51465402275837 |
10 | 13140314143531 |
11 | 4206859848593 |
12 | 1582820752837 |
13 | 74418465aa53 |
14 | 335dcd269591 |
15 | 17bc22c5cd71 |
hex | bf377c1bb2b |
13140314143531 has 2 divisors, whose sum is σ = 13140314143532. Its totient is φ = 13140314143530.
The previous prime is 13140314143487. The next prime is 13140314143613. The reversal of 13140314143531 is 13534141304131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13140314143531 - 213 = 13140314135339 is a prime.
It is a super-2 number, since 2×131403141435312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13140314743531) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6570157071765 + 6570157071766.
It is an arithmetic number, because the mean of its divisors is an integer number (6570157071766).
Almost surely, 213140314143531 is an apocalyptic number.
13140314143531 is a deficient number, since it is larger than the sum of its proper divisors (1).
13140314143531 is an equidigital number, since it uses as much as digits as its factorization.
13140314143531 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 25920, while the sum is 34.
Adding to 13140314143531 its reverse (13534141304131), we get a palindrome (26674455447662).
The spelling of 13140314143531 in words is "thirteen trillion, one hundred forty billion, three hundred fourteen million, one hundred forty-three thousand, five hundred thirty-one".
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