Base | Representation |
---|---|
bin | 11101101110001101110… |
… | …10010001001001001011 |
3 | 10121122000021112101202022 |
4 | 32313012322101021023 |
5 | 113213002032144431 |
6 | 2101053031555055 |
7 | 133532250164462 |
oct | 16670672211113 |
9 | 3548007471668 |
10 | 1021244412491 |
11 | 36411a935496 |
12 | 145b1082ba8b |
13 | 753c2717a14 |
14 | 375ddadd9d9 |
15 | 1b87189207b |
hex | edc6e9124b |
1021244412491 has 2 divisors, whose sum is σ = 1021244412492. Its totient is φ = 1021244412490.
The previous prime is 1021244412479. The next prime is 1021244412493. The reversal of 1021244412491 is 1942144421201.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1021244412491 is a prime.
Together with 1021244412493, 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 (1021244412493) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 510622206245 + 510622206246.
It is an arithmetic number, because the mean of its divisors is an integer number (510622206246).
Almost surely, 21021244412491 is an apocalyptic number.
1021244412491 is a deficient number, since it is larger than the sum of its proper divisors (1).
1021244412491 is an equidigital number, since it uses as much as digits as its factorization.
1021244412491 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 18432, while the sum is 35.
Adding to 1021244412491 its reverse (1942144421201), we get a palindrome (2963388833692).
The spelling of 1021244412491 in words is "one trillion, twenty-one billion, two hundred forty-four million, four hundred twelve thousand, four hundred ninety-one".
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