Base | Representation |
---|---|
bin | 11001111110100111010110… |
… | …111100000110100000010111 |
3 | 112222112120121212121110210022 |
4 | 121332213112330012200113 |
5 | 104433414223431323112 |
6 | 1042555324410215355 |
7 | 33031403254442243 |
oct | 3176472674064027 |
9 | 488476555543708 |
10 | 114254031120407 |
11 | 3344a953a75482 |
12 | 10993243a2455b |
13 | 4b9a155355409 |
14 | 202dcd31c7423 |
15 | d32022357d72 |
hex | 67e9d6f06817 |
114254031120407 has 2 divisors, whose sum is σ = 114254031120408. Its totient is φ = 114254031120406.
The previous prime is 114254031120287. The next prime is 114254031120521. The reversal of 114254031120407 is 704021130452411.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 114254031120407 - 210 = 114254031119383 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (114254031140407) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57127015560203 + 57127015560204.
It is an arithmetic number, because the mean of its divisors is an integer number (57127015560204).
Almost surely, 2114254031120407 is an apocalyptic number.
114254031120407 is a deficient number, since it is larger than the sum of its proper divisors (1).
114254031120407 is an equidigital number, since it uses as much as digits as its factorization.
114254031120407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 26880, while the sum is 35.
Adding to 114254031120407 its reverse (704021130452411), we get a palindrome (818275161572818).
The spelling of 114254031120407 in words is "one hundred fourteen trillion, two hundred fifty-four billion, thirty-one million, one hundred twenty thousand, four hundred seven".
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