Base | Representation |
---|---|
bin | 1100000111111110100110… |
… | …1100010010100011110011 |
3 | 1202012110011112200020021211 |
4 | 3001333221230102203303 |
5 | 3221404301240034141 |
6 | 44204134244300551 |
7 | 2544101445131554 |
oct | 301775154224363 |
9 | 52173145606254 |
10 | 13331204221171 |
11 | 427a80626a231 |
12 | 15b38153a1757 |
13 | 7591864c0a05 |
14 | 34133b4c032b |
15 | 181b96540d81 |
hex | c1fe9b128f3 |
13331204221171 has 2 divisors, whose sum is σ = 13331204221172. Its totient is φ = 13331204221170.
The previous prime is 13331204221169. The next prime is 13331204221213. The reversal of 13331204221171 is 17112240213331.
It is a happy number.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13331204221171 - 21 = 13331204221169 is a prime.
It is a super-2 number, since 2×133312042211712 (a number of 27 digits) contains 22 as substring.
Together with 13331204221169, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (13331204221771) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6665602110585 + 6665602110586.
It is an arithmetic number, because the mean of its divisors is an integer number (6665602110586).
Almost surely, 213331204221171 is an apocalyptic number.
13331204221171 is a deficient number, since it is larger than the sum of its proper divisors (1).
13331204221171 is an equidigital number, since it uses as much as digits as its factorization.
13331204221171 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048, while the sum is 31.
The spelling of 13331204221171 in words is "thirteen trillion, three hundred thirty-one billion, two hundred four million, two hundred twenty-one thousand, one hundred seventy-one".
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