Base | Representation |
---|---|
bin | 10011110110100001000… |
… | …100001111101101011011 |
3 | 11211102020220220210001022 |
4 | 103312201010033231123 |
5 | 134322343430040111 |
6 | 2522412531513055 |
7 | 200363214063221 |
oct | 23664104175533 |
9 | 4742226823038 |
10 | 1364206877531 |
11 | 486614243643 |
12 | 1a048615878b |
13 | 9b84b29a874 |
14 | 4a056a44b11 |
15 | 25745b652db |
hex | 13da110fb5b |
1364206877531 has 2 divisors, whose sum is σ = 1364206877532. Its totient is φ = 1364206877530.
The previous prime is 1364206877489. The next prime is 1364206877533. The reversal of 1364206877531 is 1357786024631.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1364206877531 is a prime.
It is a super-2 number, since 2×13642068775312 (a number of 25 digits) contains 22 as substring.
Together with 1364206877533, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1364206877533) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 682103438765 + 682103438766.
It is an arithmetic number, because the mean of its divisors is an integer number (682103438766).
Almost surely, 21364206877531 is an apocalyptic number.
1364206877531 is a deficient number, since it is larger than the sum of its proper divisors (1).
1364206877531 is an equidigital number, since it uses as much as digits as its factorization.
1364206877531 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5080320, while the sum is 53.
The spelling of 1364206877531 in words is "one trillion, three hundred sixty-four billion, two hundred six million, eight hundred seventy-seven thousand, five hundred thirty-one".
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