Base | Representation |
---|---|
bin | 111001110101110110110… |
… | …111101001010010010001 |
3 | 112001222202100220212112121 |
4 | 321311312313221102101 |
5 | 1010110432244003223 |
6 | 12242003231254241 |
7 | 560113134303661 |
oct | 71656667512221 |
9 | 15058670825477 |
10 | 3974839047313 |
11 | 12a27a10224a2 |
12 | 54242567a381 |
13 | 22aa97129c54 |
14 | da55185a1a1 |
15 | 6d5dc0e315d |
hex | 39d76de9491 |
3974839047313 has 2 divisors, whose sum is σ = 3974839047314. Its totient is φ = 3974839047312.
The previous prime is 3974839047301. The next prime is 3974839047329. The reversal of 3974839047313 is 3137409384793.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2511109283904 + 1463729763409 = 1584648^2 + 1209847^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-3974839047313 is a prime.
It is not a weakly prime, because it can be changed into another prime (3974839042313) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1987419523656 + 1987419523657.
It is an arithmetic number, because the mean of its divisors is an integer number (1987419523657).
Almost surely, 23974839047313 is an apocalyptic number.
It is an amenable number.
3974839047313 is a deficient number, since it is larger than the sum of its proper divisors (1).
3974839047313 is an equidigital number, since it uses as much as digits as its factorization.
3974839047313 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 41150592, while the sum is 61.
The spelling of 3974839047313 in words is "three trillion, nine hundred seventy-four billion, eight hundred thirty-nine million, forty-seven thousand, three hundred thirteen".
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