Base | Representation |
---|---|
bin | 1011111011110100110011… |
… | …0100110011011111000011 |
3 | 1201110111020121001001022211 |
4 | 2332331030310303133003 |
5 | 3204444200441000232 |
6 | 43524204225242551 |
7 | 2523030436420021 |
oct | 276751464633703 |
9 | 51414217031284 |
10 | 13122414000067 |
11 | 41aa203669899 |
12 | 157b265a93457 |
13 | 742591020898 |
14 | 3351b1bd3711 |
15 | 17b52652ba47 |
hex | bef4cd337c3 |
13122414000067 has 2 divisors, whose sum is σ = 13122414000068. Its totient is φ = 13122414000066.
The previous prime is 13122414000041. The next prime is 13122414000109. The reversal of 13122414000067 is 76000041422131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13122414000067 - 235 = 13088054261699 is a prime.
It is not a weakly prime, because it can be changed into another prime (13122414000467) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6561207000033 + 6561207000034.
It is an arithmetic number, because the mean of its divisors is an integer number (6561207000034).
Almost surely, 213122414000067 is an apocalyptic number.
13122414000067 is a deficient number, since it is larger than the sum of its proper divisors (1).
13122414000067 is an equidigital number, since it uses as much as digits as its factorization.
13122414000067 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 8064, while the sum is 31.
Adding to 13122414000067 its reverse (76000041422131), we get a palindrome (89122455422198).
The spelling of 13122414000067 in words is "thirteen trillion, one hundred twenty-two billion, four hundred fourteen million, sixty-seven", and thus it is an aban number.
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