Base | Representation |
---|---|
bin | 110111110011000011110111… |
… | …100010001000000000100011 |
3 | 1012011220010220210202112202221 |
4 | 313303003313202020000203 |
5 | 224131124034002423212 |
6 | 2225531503202251511 |
7 | 102455444365634113 |
oct | 6763036742100043 |
9 | 1164803823675687 |
10 | 245401404342307 |
11 | 71213173171454 |
12 | 23634529956b97 |
13 | a6c133a197cc5 |
14 | 44856c436b043 |
15 | 1d586c1327807 |
hex | df30f7888023 |
245401404342307 has 2 divisors, whose sum is σ = 245401404342308. Its totient is φ = 245401404342306.
The previous prime is 245401404342217. The next prime is 245401404342317. The reversal of 245401404342307 is 703243404104542.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 245401404342307 - 231 = 245399256858659 is a prime.
It is a super-3 number, since 3×2454014043423073 (a number of 44 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (245401404342317) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 122700702171153 + 122700702171154.
It is an arithmetic number, because the mean of its divisors is an integer number (122700702171154).
Almost surely, 2245401404342307 is an apocalyptic number.
245401404342307 is a deficient number, since it is larger than the sum of its proper divisors (1).
245401404342307 is an equidigital number, since it uses as much as digits as its factorization.
245401404342307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1290240, while the sum is 43.
Adding to 245401404342307 its reverse (703243404104542), we get a palindrome (948644808446849).
The spelling of 245401404342307 in words is "two hundred forty-five trillion, four hundred one billion, four hundred four million, three hundred forty-two thousand, three hundred seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.035 sec. • engine limits •