Base | Representation |
---|---|
bin | 1010111011001110001100… |
… | …1100111011010111100011 |
3 | 1120112101110121000020102002 |
4 | 2232303203030323113203 |
5 | 3033303140314022142 |
6 | 41314252421514215 |
7 | 2346606643320044 |
oct | 256634314732743 |
9 | 46471417006362 |
10 | 12012540376547 |
11 | 3911542970367 |
12 | 142013b57296b |
13 | 691a157a5539 |
14 | 2d75a52a28cb |
15 | 15c718d32a32 |
hex | aece333b5e3 |
12012540376547 has 2 divisors, whose sum is σ = 12012540376548. Its totient is φ = 12012540376546.
The previous prime is 12012540376481. The next prime is 12012540376549. The reversal of 12012540376547 is 74567304521021.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 12012540376547 - 224 = 12012523599331 is a prime.
It is a super-2 number, since 2×120125403765472 (a number of 27 digits) contains 22 as substring.
Together with 12012540376549, 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 (12012540376549) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6006270188273 + 6006270188274.
It is an arithmetic number, because the mean of its divisors is an integer number (6006270188274).
Almost surely, 212012540376547 is an apocalyptic number.
12012540376547 is a deficient number, since it is larger than the sum of its proper divisors (1).
12012540376547 is an equidigital number, since it uses as much as digits as its factorization.
12012540376547 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1411200, while the sum is 47.
Adding to 12012540376547 its reverse (74567304521021), we get a palindrome (86579844897568).
The spelling of 12012540376547 in words is "twelve trillion, twelve billion, five hundred forty million, three hundred seventy-six thousand, five hundred forty-seven".
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