Base | Representation |
---|---|
bin | 110011010010001110… |
… | …0111110011001101011 |
3 | 101112021101010100021122 |
4 | 1212210130332121223 |
5 | 3301023102211301 |
6 | 122332225325455 |
7 | 10646130051332 |
oct | 1464434763153 |
9 | 345241110248 |
10 | 110133241451 |
11 | 42786386412 |
12 | 1941749688b |
13 | a501cabbbc |
14 | 548ab88d19 |
15 | 2ce8b6681b |
hex | 19a473e66b |
110133241451 has 2 divisors, whose sum is σ = 110133241452. Its totient is φ = 110133241450.
The previous prime is 110133241447. The next prime is 110133241453. The reversal of 110133241451 is 154142331011.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 110133241451 - 22 = 110133241447 is a prime.
Together with 110133241453, it forms a pair of twin primes.
It is a Chen prime.
It is a self number, because there is not a number n which added to its sum of digits gives 110133241451.
It is not a weakly prime, because it can be changed into another prime (110133241453) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55066620725 + 55066620726.
It is an arithmetic number, because the mean of its divisors is an integer number (55066620726).
Almost surely, 2110133241451 is an apocalyptic number.
110133241451 is a deficient number, since it is larger than the sum of its proper divisors (1).
110133241451 is an equidigital number, since it uses as much as digits as its factorization.
110133241451 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1440, while the sum is 26.
Adding to 110133241451 its reverse (154142331011), we get a palindrome (264275572462).
The spelling of 110133241451 in words is "one hundred ten billion, one hundred thirty-three million, two hundred forty-one thousand, four hundred fifty-one".
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