Base | Representation |
---|---|
bin | 11101101110010110001… |
… | …00111010101110111011 |
3 | 10121122012011001201102111 |
4 | 32313023010322232323 |
5 | 113213122431010042 |
6 | 2101104010113151 |
7 | 133534056265462 |
oct | 16671304725673 |
9 | 3548164051374 |
10 | 1021314313147 |
11 | 364156338927 |
12 | 145b3011b7b7 |
13 | 7540405025b |
14 | 376090d79d9 |
15 | 1b877a9d517 |
hex | edcb13abbb |
1021314313147 has 2 divisors, whose sum is σ = 1021314313148. Its totient is φ = 1021314313146.
The previous prime is 1021314313109. The next prime is 1021314313153. The reversal of 1021314313147 is 7413134131201.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1021314313147 - 215 = 1021314280379 is a prime.
It is not a weakly prime, because it can be changed into another prime (1021374313147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 510657156573 + 510657156574.
It is an arithmetic number, because the mean of its divisors is an integer number (510657156574).
Almost surely, 21021314313147 is an apocalyptic number.
1021314313147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1021314313147 is an equidigital number, since it uses as much as digits as its factorization.
1021314313147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6048, while the sum is 31.
Adding to 1021314313147 its reverse (7413134131201), we get a palindrome (8434448444348).
The spelling of 1021314313147 in words is "one trillion, twenty-one billion, three hundred fourteen million, three hundred thirteen thousand, one hundred forty-seven".
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