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111131030999993 is a prime number
BaseRepresentation
bin11001010001001010110101…
…101000000101001110111001
3112120111000121121220200022212
4121101022311220011032321
5104031232322413444433
61032204520231224505
732256644355426044
oct3121126550051671
9476430547820285
10111131030999993
1132456450901824
1210569b31518135
134a017c3c13b94
141d62ab074dc5b
15ccab8e593d48
hex6512b5a053b9

111131030999993 has 2 divisors, whose sum is σ = 111131030999994. Its totient is φ = 111131030999992.

The previous prime is 111131030999881. The next prime is 111131031000037. The reversal of 111131030999993 is 399999030131111.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 110317378291264 + 813652708729 = 10503208^2 + 902027^2 .

It is an emirp because it is prime and its reverse (399999030131111) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 111131030999993 - 216 = 111131030934457 is a prime.

It is a super-2 number, since 2×1111310309999932 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (111131080999993) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55565515499996 + 55565515499997.

It is an arithmetic number, because the mean of its divisors is an integer number (55565515499997).

Almost surely, 2111131030999993 is an apocalyptic number.

It is an amenable number.

111131030999993 is a deficient number, since it is larger than the sum of its proper divisors (1).

111131030999993 is an equidigital number, since it uses as much as digits as its factorization.

111131030999993 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1594323, while the sum is 59.

The spelling of 111131030999993 in words is "one hundred eleven trillion, one hundred thirty-one billion, thirty million, nine hundred ninety-nine thousand, nine hundred ninety-three".