Base | Representation |
---|---|
bin | 100100101000010010001… |
… | …0101100011010100000001 |
3 | 122211021102210212210001011 |
4 | 1021100210111203110001 |
5 | 1124440232143102403 |
6 | 14412421454440521 |
7 | 1026500633000566 |
oct | 111204425432401 |
9 | 18737383783034 |
10 | 5034311300353 |
11 | 1671048537335 |
12 | 693824803141 |
13 | 2a696c70425a |
14 | 13593a61a86d |
15 | 8ae49ab2a6d |
hex | 49424563501 |
5034311300353 has 2 divisors, whose sum is σ = 5034311300354. Its totient is φ = 5034311300352.
The previous prime is 5034311300321. The next prime is 5034311300401. The reversal of 5034311300353 is 3530031134305.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 4412717818609 + 621593481744 = 2100647^2 + 788412^2 .
It is a cyclic number.
It is not a de Polignac number, because 5034311300353 - 25 = 5034311300321 is a prime.
It is not a weakly prime, because it can be changed into another prime (5034311300453) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2517155650176 + 2517155650177.
It is an arithmetic number, because the mean of its divisors is an integer number (2517155650177).
Almost surely, 25034311300353 is an apocalyptic number.
It is an amenable number.
5034311300353 is a deficient number, since it is larger than the sum of its proper divisors (1).
5034311300353 is an equidigital number, since it uses as much as digits as its factorization.
5034311300353 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 24300, while the sum is 31.
Adding to 5034311300353 its reverse (3530031134305), we get a palindrome (8564342434658).
The spelling of 5034311300353 in words is "five trillion, thirty-four billion, three hundred eleven million, three hundred thousand, three hundred fifty-three".
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