Base | Representation |
---|---|
bin | 11110010111001011010001… |
… | …010100011111001001000111 |
3 | 122111210201201012020120020011 |
4 | 132113023101110133021013 |
5 | 120000310211012304343 |
6 | 1152000405440051051 |
7 | 40061342056003201 |
oct | 3627132124371107 |
9 | 574721635216204 |
10 | 133534045041223 |
11 | 3960351a222802 |
12 | 12b87961603a87 |
13 | 596828b659958 |
14 | 24d9121a3d771 |
15 | 10687d871a29d |
hex | 7972d151f247 |
133534045041223 has 2 divisors, whose sum is σ = 133534045041224. Its totient is φ = 133534045041222.
The previous prime is 133534045041149. The next prime is 133534045041227. The reversal of 133534045041223 is 322140540435331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133534045041223 - 217 = 133534044910151 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (133534045041227) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66767022520611 + 66767022520612.
It is an arithmetic number, because the mean of its divisors is an integer number (66767022520612).
Almost surely, 2133534045041223 is an apocalyptic number.
133534045041223 is a deficient number, since it is larger than the sum of its proper divisors (1).
133534045041223 is an equidigital number, since it uses as much as digits as its factorization.
133534045041223 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 518400, while the sum is 40.
Adding to 133534045041223 its reverse (322140540435331), we get a palindrome (455674585476554).
The spelling of 133534045041223 in words is "one hundred thirty-three trillion, five hundred thirty-four billion, forty-five million, forty-one thousand, two hundred twenty-three".
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