Base | Representation |
---|---|
bin | 101101100001011101010100… |
… | …100111011100111010011111 |
3 | 222020220000102010201222210111 |
4 | 231201131110213130322133 |
5 | 202220230240423144411 |
6 | 1545451451041150451 |
7 | 60112535103640543 |
oct | 5541352447347237 |
9 | 866800363658714 |
10 | 200211320131231 |
11 | 58880164796702 |
12 | 1a55636602b427 |
13 | 8793b00931556 |
14 | 37623cb309a23 |
15 | 1822e5123b721 |
hex | b617549dce9f |
200211320131231 has 2 divisors, whose sum is σ = 200211320131232. Its totient is φ = 200211320131230.
The previous prime is 200211320131099. The next prime is 200211320131349. The reversal of 200211320131231 is 132131023112002.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 200211320131231 - 29 = 200211320130719 is a prime.
It is a super-2 number, since 2×2002113201312312 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (200211320131631) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100105660065615 + 100105660065616.
It is an arithmetic number, because the mean of its divisors is an integer number (100105660065616).
Almost surely, 2200211320131231 is an apocalyptic number.
200211320131231 is a deficient number, since it is larger than the sum of its proper divisors (1).
200211320131231 is an equidigital number, since it uses as much as digits as its factorization.
200211320131231 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 432, while the sum is 22.
Adding to 200211320131231 its reverse (132131023112002), we get a palindrome (332342343243233).
The spelling of 200211320131231 in words is "two hundred trillion, two hundred eleven billion, three hundred twenty million, one hundred thirty-one thousand, two hundred thirty-one".
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