Base | Representation |
---|---|
bin | 11111000111001010100101… |
… | …101101111101000101000011 |
3 | 122221110222220210002002122001 |
4 | 133013022211231331011003 |
5 | 120413323111142401032 |
6 | 1203003403335124431 |
7 | 40551526522035433 |
oct | 3707124555750503 |
9 | 587428823062561 |
10 | 136831848403267 |
11 | 3a665074252106 |
12 | 1341ab19213717 |
13 | 5b47259b93184 |
14 | 25b09a80342c3 |
15 | 10c449c8673e7 |
hex | 7c72a5b7d143 |
136831848403267 has 2 divisors, whose sum is σ = 136831848403268. Its totient is φ = 136831848403266.
The previous prime is 136831848403213. The next prime is 136831848403291. The reversal of 136831848403267 is 762304848138631.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-136831848403267 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 136831848403199 and 136831848403208.
It is not a weakly prime, because it can be changed into another prime (136831848401267) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 68415924201633 + 68415924201634.
It is an arithmetic number, because the mean of its divisors is an integer number (68415924201634).
Almost surely, 2136831848403267 is an apocalyptic number.
136831848403267 is a deficient number, since it is larger than the sum of its proper divisors (1).
136831848403267 is an equidigital number, since it uses as much as digits as its factorization.
136831848403267 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 111476736, while the sum is 64.
The spelling of 136831848403267 in words is "one hundred thirty-six trillion, eight hundred thirty-one billion, eight hundred forty-eight million, four hundred three thousand, two hundred sixty-seven".
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