Base | Representation |
---|---|
bin | 11000111011101111111… |
… | …000001110111001000011 |
3 | 20001210122021210121211222 |
4 | 120323233320032321003 |
5 | 211033041344320431 |
6 | 3351045052425255 |
7 | 234535116162251 |
oct | 30735770167103 |
9 | 6053567717758 |
10 | 1713421479491 |
11 | 6007268357a7 |
12 | 2380a547822b |
13 | c5762229707 |
14 | 5cd0406d5d1 |
15 | 2e883c7857b |
hex | 18eefe0ee43 |
1713421479491 has 2 divisors, whose sum is σ = 1713421479492. Its totient is φ = 1713421479490.
The previous prime is 1713421479487. The next prime is 1713421479503. The reversal of 1713421479491 is 1949741243171.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1713421479491 - 22 = 1713421479487 is a prime.
It is a super-2 number, since 2×17134214794912 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1713421479401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 856710739745 + 856710739746.
It is an arithmetic number, because the mean of its divisors is an integer number (856710739746).
Almost surely, 21713421479491 is an apocalyptic number.
1713421479491 is a deficient number, since it is larger than the sum of its proper divisors (1).
1713421479491 is an equidigital number, since it uses as much as digits as its factorization.
1713421479491 is an evil number, because the sum of its binary digits is even.
The product of its digits is 1524096, while the sum is 53.
It can be divided in two parts, 1713421 and 479491, that added together give a palindrome (2192912).
The spelling of 1713421479491 in words is "one trillion, seven hundred thirteen billion, four hundred twenty-one million, four hundred seventy-nine thousand, four hundred ninety-one".
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