Base | Representation |
---|---|
bin | 111100010000100111111101… |
… | …0101001011011001110000011 |
3 | 2120111202022202201100120011112 |
4 | 1320201033322221123032003 |
5 | 1023433321222004130201 |
6 | 5115153444302433535 |
7 | 216434610246522113 |
oct | 17041177251331603 |
9 | 2514668681316145 |
10 | 530050414130051 |
11 | 143988133432a11 |
12 | 4b5473a32928ab |
13 | 1999b7025c0909 |
14 | 94c6811139043 |
15 | 4142c635c20bb |
hex | 1e213faa5b383 |
530050414130051 has 2 divisors, whose sum is σ = 530050414130052. Its totient is φ = 530050414130050.
The previous prime is 530050414130033. The next prime is 530050414130123. The reversal of 530050414130051 is 150031414050035.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 530050414130051 - 26 = 530050414129987 is a prime.
It is not a weakly prime, because it can be changed into another prime (530050414130651) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265025207065025 + 265025207065026.
It is an arithmetic number, because the mean of its divisors is an integer number (265025207065026).
Almost surely, 2530050414130051 is an apocalyptic number.
530050414130051 is a deficient number, since it is larger than the sum of its proper divisors (1).
530050414130051 is an equidigital number, since it uses as much as digits as its factorization.
530050414130051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 18000, while the sum is 32.
Adding to 530050414130051 its reverse (150031414050035), we get a palindrome (680081828180086).
The spelling of 530050414130051 in words is "five hundred thirty trillion, fifty billion, four hundred fourteen million, one hundred thirty thousand, fifty-one".
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