Base | Representation |
---|---|
bin | 11001111110101000010010… |
… | …101011011111111100111011 |
3 | 112222112200012201120121111012 |
4 | 121332220102223133330323 |
5 | 104433423302022314042 |
6 | 1043000012052424135 |
7 | 33031440144622442 |
oct | 3176502253377473 |
9 | 488480181517435 |
10 | 114255033401147 |
11 | 33450318806871 |
12 | 1099348361804b |
13 | 4b9a284bb0742 |
14 | 202dd8a369d59 |
15 | d32080339e82 |
hex | 67ea12adff3b |
114255033401147 has 2 divisors, whose sum is σ = 114255033401148. Its totient is φ = 114255033401146.
The previous prime is 114255033401143. The next prime is 114255033401213. The reversal of 114255033401147 is 741104330552411.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 114255033401147 - 22 = 114255033401143 is a prime.
It is not a weakly prime, because it can be changed into another prime (114255033401141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57127516700573 + 57127516700574.
It is an arithmetic number, because the mean of its divisors is an integer number (57127516700574).
Almost surely, 2114255033401147 is an apocalyptic number.
114255033401147 is a deficient number, since it is larger than the sum of its proper divisors (1).
114255033401147 is an equidigital number, since it uses as much as digits as its factorization.
114255033401147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 201600, while the sum is 41.
Adding to 114255033401147 its reverse (741104330552411), we get a palindrome (855359363953558).
The spelling of 114255033401147 in words is "one hundred fourteen trillion, two hundred fifty-five billion, thirty-three million, four hundred one thousand, one hundred forty-seven".
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