Base | Representation |
---|---|
bin | 11111000001101101100… |
… | …001101111101000111111 |
3 | 21112211102121002111112222 |
4 | 133001231201233220333 |
5 | 234413111140404442 |
6 | 4311254152315555 |
7 | 310020310241561 |
oct | 37015541575077 |
9 | 7484377074488 |
10 | 2132141341247 |
11 | 752266225325 |
12 | 2a52819a2bbb |
13 | 1260a1c6cc18 |
14 | 752a64c7331 |
15 | 3a6ddd228d2 |
hex | 1f06d86fa3f |
2132141341247 has 2 divisors, whose sum is σ = 2132141341248. Its totient is φ = 2132141341246.
The previous prime is 2132141341223. The next prime is 2132141341273. The reversal of 2132141341247 is 7421431412312.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2132141341247 - 216 = 2132141275711 is a prime.
It is a super-2 number, since 2×21321413412472 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2132141341207) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1066070670623 + 1066070670624.
It is an arithmetic number, because the mean of its divisors is an integer number (1066070670624).
Almost surely, 22132141341247 is an apocalyptic number.
2132141341247 is a deficient number, since it is larger than the sum of its proper divisors (1).
2132141341247 is an equidigital number, since it uses as much as digits as its factorization.
2132141341247 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 32256, while the sum is 35.
Adding to 2132141341247 its reverse (7421431412312), we get a palindrome (9553572753559).
The spelling of 2132141341247 in words is "two trillion, one hundred thirty-two billion, one hundred forty-one million, three hundred forty-one thousand, two hundred forty-seven".
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