Base | Representation |
---|---|
bin | 100100101000010000000… |
… | …0110000010000100001111 |
3 | 122211021020211221121021111 |
4 | 1021100200012002010033 |
5 | 1124440110432420101 |
6 | 14412410433112451 |
7 | 1026466110002101 |
oct | 111204006020417 |
9 | 18737224847244 |
10 | 5034240123151 |
11 | 16710113418a8 |
12 | 6938049b8727 |
13 | 2a695aa42975 |
14 | 135930dad571 |
15 | 8ae43703251 |
hex | 4942018210f |
5034240123151 has 4 divisors (see below), whose sum is σ = 5039320083904. Its totient is φ = 5029160162400.
The previous prime is 5034240123137. The next prime is 5034240123193. The reversal of 5034240123151 is 1513210424305.
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 5034240123151 - 225 = 5034206568719 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 (5094240123151) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2539978890 + ... + 2539980871.
It is an arithmetic number, because the mean of its divisors is an integer number (1259830020976).
Almost surely, 25034240123151 is an apocalyptic number.
5034240123151 is a deficient number, since it is larger than the sum of its proper divisors (5079960753).
5034240123151 is an equidigital number, since it uses as much as digits as its factorization.
5034240123151 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 5079960752.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 5034240123151 its reverse (1513210424305), we get a palindrome (6547450547456).
The spelling of 5034240123151 in words is "five trillion, thirty-four billion, two hundred forty million, one hundred twenty-three thousand, one hundred fifty-one".
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