Base | Representation |
---|---|
bin | 100000001111011100011… |
… | …0000110111110000001001 |
3 | 120200121201201110021210211 |
4 | 1000132320300313300021 |
5 | 1040100110020214233 |
6 | 13231400440241121 |
7 | 635100353615212 |
oct | 100367060676011 |
9 | 16617651407724 |
10 | 4431211101193 |
11 | 14592a2166278 |
12 | 5b696b7701a1 |
13 | 261b276ac196 |
14 | 11468672a409 |
15 | 7a3ec9ac5cd |
hex | 407b8c37c09 |
4431211101193 has 2 divisors, whose sum is σ = 4431211101194. Its totient is φ = 4431211101192.
The previous prime is 4431211101179. The next prime is 4431211101199. The reversal of 4431211101193 is 3911011121344.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3772560366864 + 658650734329 = 1942308^2 + 811573^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-4431211101193 is a prime.
It is not a weakly prime, because it can be changed into another prime (4431211101199) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2215605550596 + 2215605550597.
It is an arithmetic number, because the mean of its divisors is an integer number (2215605550597).
Almost surely, 24431211101193 is an apocalyptic number.
It is an amenable number.
4431211101193 is a deficient number, since it is larger than the sum of its proper divisors (1).
4431211101193 is an equidigital number, since it uses as much as digits as its factorization.
4431211101193 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2592, while the sum is 31.
The spelling of 4431211101193 in words is "four trillion, four hundred thirty-one billion, two hundred eleven million, one hundred one thousand, one hundred ninety-three".
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