Base | Representation |
---|---|
bin | 111111001011001010011… |
… | …100011001101110110111 |
3 | 120101000200121002111202111 |
4 | 333023022130121232313 |
5 | 1032112002213141043 |
6 | 13122212152551451 |
7 | 625435536100030 |
oct | 77131234315667 |
9 | 16330617074674 |
10 | 4341313412023 |
11 | 1424160274895 |
12 | 5a14610b2587 |
13 | 2564cbb2c473 |
14 | 110199261a87 |
15 | 77dda5e009d |
hex | 3f2ca719bb7 |
4341313412023 has 4 divisors (see below), whose sum is σ = 4961501042320. Its totient is φ = 3721125781728.
The previous prime is 4341313411879. The next prime is 4341313412027. The reversal of 4341313412023 is 3202143131434.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 4341313412023 - 233 = 4332723477431 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (4341313412027) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 310093815138 + ... + 310093815151.
It is an arithmetic number, because the mean of its divisors is an integer number (1240375260580).
Almost surely, 24341313412023 is an apocalyptic number.
4341313412023 is a deficient number, since it is larger than the sum of its proper divisors (620187630297).
4341313412023 is an equidigital number, since it uses as much as digits as its factorization.
4341313412023 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 620187630296.
The product of its (nonzero) digits is 20736, while the sum is 31.
Adding to 4341313412023 its reverse (3202143131434), we get a palindrome (7543456543457).
The spelling of 4341313412023 in words is "four trillion, three hundred forty-one billion, three hundred thirteen million, four hundred twelve thousand, twenty-three".
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