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2031110001433 is a prime number
BaseRepresentation
bin11101100011100111100…
…101110001001100011001
321012011122112210200001101
4131203213211301030121
5231234203130021213
64153025024151401
7266512541103502
oct35434745611431
97164575720041
102031110001433
11713430710377
12289786768561
131196c089c546
14704404bbba9
1537c792b5bdd
hex1d8e7971319

2031110001433 has 2 divisors, whose sum is σ = 2031110001434. Its totient is φ = 2031110001432.

The previous prime is 2031110001413. The next prime is 2031110001451. The reversal of 2031110001433 is 3341000111302.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1146584799369 + 884525202064 = 1070787^2 + 940492^2 .

It is a cyclic number.

It is not a de Polignac number, because 2031110001433 - 221 = 2031107904281 is a prime.

It is a super-2 number, since 2×20311100014332 (a number of 25 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 2031110001433.

It is not a weakly prime, because it can be changed into another prime (2031110001413) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1015555000716 + 1015555000717.

It is an arithmetic number, because the mean of its divisors is an integer number (1015555000717).

Almost surely, 22031110001433 is an apocalyptic number.

It is an amenable number.

2031110001433 is a deficient number, since it is larger than the sum of its proper divisors (1).

2031110001433 is an equidigital number, since it uses as much as digits as its factorization.

2031110001433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 216, while the sum is 19.

Adding to 2031110001433 its reverse (3341000111302), we get a palindrome (5372110112735).

The spelling of 2031110001433 in words is "two trillion, thirty-one billion, one hundred ten million, one thousand, four hundred thirty-three".