Base | Representation |
---|---|
bin | 100101001110110000000… |
… | …1100101110111110110001 |
3 | 200010011122221201112210121 |
4 | 1022131200030232332301 |
5 | 1132313423010003213 |
6 | 14514403001530241 |
7 | 1035454022240023 |
oct | 112354014567661 |
9 | 20104587645717 |
10 | 5116920000433 |
11 | 16a3089a4039a |
12 | 6a783a1a3981 |
13 | 2b16a6136c87 |
14 | 139935a20013 |
15 | 8d1820ce88d |
hex | 4a76032efb1 |
5116920000433 has 2 divisors, whose sum is σ = 5116920000434. Its totient is φ = 5116920000432.
The previous prime is 5116920000419. The next prime is 5116920000539. The reversal of 5116920000433 is 3340000296115.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 4927654108224 + 189265892209 = 2219832^2 + 435047^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5116920000433 is a prime.
It is not a weakly prime, because it can be changed into another prime (5116920000733) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2558460000216 + 2558460000217.
It is an arithmetic number, because the mean of its divisors is an integer number (2558460000217).
Almost surely, 25116920000433 is an apocalyptic number.
It is an amenable number.
5116920000433 is a deficient number, since it is larger than the sum of its proper divisors (1).
5116920000433 is an equidigital number, since it uses as much as digits as its factorization.
5116920000433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 19440, while the sum is 34.
Adding to 5116920000433 its reverse (3340000296115), we get a palindrome (8456920296548).
The spelling of 5116920000433 in words is "five trillion, one hundred sixteen billion, nine hundred twenty million, four hundred thirty-three", and thus it is an aban number.
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