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101111110099981 is a prime number
BaseRepresentation
bin10110111111010111000011…
…110011001110010000001101
3111021000010110002222212000221
4112333113003303032100031
5101223101023141144411
6555013445220151341
730204015365151652
oct2677270363162015
9437003402885027
10101111110099981
112a242a87765783
12b410016238551
1344559768b7577
141ad7b37a49c29
15ba51ea80c071
hex5bf5c3cce40d

101111110099981 has 2 divisors, whose sum is σ = 101111110099982. Its totient is φ = 101111110099980.

The previous prime is 101111110099969. The next prime is 101111110100009. The reversal of 101111110099981 is 189990011111101.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 90354245085225 + 10756865014756 = 9505485^2 + 3279766^2 .

It is a cyclic number.

It is not a de Polignac number, because 101111110099981 - 29 = 101111110099469 is a prime.

It is a super-3 number, since 3×1011111100999813 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (101111110009981) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50555555049990 + 50555555049991.

It is an arithmetic number, because the mean of its divisors is an integer number (50555555049991).

Almost surely, 2101111110099981 is an apocalyptic number.

It is an amenable number.

101111110099981 is a deficient number, since it is larger than the sum of its proper divisors (1).

101111110099981 is an equidigital number, since it uses as much as digits as its factorization.

101111110099981 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 5832, while the sum is 43.

The spelling of 101111110099981 in words is "one hundred one trillion, one hundred eleven billion, one hundred ten million, ninety-nine thousand, nine hundred eighty-one".