Base | Representation |
---|---|
bin | 11101010010001000010… |
… | …110100100011001010011 |
3 | 21010101011220021012212211 |
4 | 131102020112210121103 |
5 | 230432224032442042 |
6 | 4140241441132551 |
7 | 265246320266623 |
oct | 35221026443123 |
9 | 7111156235784 |
10 | 2012332312147 |
11 | 7064751464a3 |
12 | 28600602a157 |
13 | 1179ba4cb1b1 |
14 | 6d57c692883 |
15 | 3752a9e1017 |
hex | 1d4885a4653 |
2012332312147 has 2 divisors, whose sum is σ = 2012332312148. Its totient is φ = 2012332312146.
The previous prime is 2012332312097. The next prime is 2012332312357. The reversal of 2012332312147 is 7412132332102.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2012332312147 - 211 = 2012332310099 is a prime.
It is not a weakly prime, because it can be changed into another prime (2012332316147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006166156073 + 1006166156074.
It is an arithmetic number, because the mean of its divisors is an integer number (1006166156074).
Almost surely, 22012332312147 is an apocalyptic number.
2012332312147 is a deficient number, since it is larger than the sum of its proper divisors (1).
2012332312147 is an equidigital number, since it uses as much as digits as its factorization.
2012332312147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12096, while the sum is 31.
Adding to 2012332312147 its reverse (7412132332102), we get a palindrome (9424464644249).
The spelling of 2012332312147 in words is "two trillion, twelve billion, three hundred thirty-two million, three hundred twelve thousand, one hundred forty-seven".
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