Base | Representation |
---|---|
bin | 10011100100101111000… |
… | …010110001100101010011 |
3 | 11202120222022100010120102 |
4 | 103210233002301211103 |
5 | 134014243142121204 |
6 | 2505534205403015 |
7 | 166116115344116 |
oct | 23445702614523 |
9 | 4676868303512 |
10 | 1345114020179 |
11 | 479506885296 |
12 | 198837b6946b |
13 | 99ac781a737 |
14 | 4916501857d |
15 | 24ec988741e |
hex | 1392f0b1953 |
1345114020179 has 2 divisors, whose sum is σ = 1345114020180. Its totient is φ = 1345114020178.
The previous prime is 1345114020143. The next prime is 1345114020197. The reversal of 1345114020179 is 9710204115431.
Together with next prime (1345114020197) it forms an Ormiston pair, because they use the same digits, order apart.
It is a strong prime.
It is an emirp because it is prime and its reverse (9710204115431) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1345114020179 - 212 = 1345114016083 is a prime.
It is a super-2 number, since 2×13451140201792 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1345114020109) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 672557010089 + 672557010090.
It is an arithmetic number, because the mean of its divisors is an integer number (672557010090).
Almost surely, 21345114020179 is an apocalyptic number.
1345114020179 is a deficient number, since it is larger than the sum of its proper divisors (1).
1345114020179 is an equidigital number, since it uses as much as digits as its factorization.
1345114020179 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 30240, while the sum is 38.
The spelling of 1345114020179 in words is "one trillion, three hundred forty-five billion, one hundred fourteen million, twenty thousand, one hundred seventy-nine".
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