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15220999931 is a prime number
BaseRepresentation
bin11100010110011111…
…00000011011111011
31110021202222220121012
432023033200123323
5222133033444211
610554210504135
71046122236044
oct161317403373
943252886535
1015220999931
116500948595
122b4959504b
131587576947
14a457118cb
155e141ad8b
hex38b3e06fb

15220999931 has 2 divisors, whose sum is σ = 15220999932. Its totient is φ = 15220999930.

The previous prime is 15220999877. The next prime is 15220999933. The reversal of 15220999931 is 13999902251.

It is a happy number.

It is a strong prime.

It is an emirp because it is prime and its reverse (13999902251) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-15220999931 is a prime.

It is a super-2 number, since 2×152209999312 (a number of 21 digits) contains 22 as substring.

Together with 15220999933, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 15220999931.

It is not a weakly prime, because it can be changed into another prime (15220999933) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7610499965 + 7610499966.

It is an arithmetic number, because the mean of its divisors is an integer number (7610499966).

Almost surely, 215220999931 is an apocalyptic number.

15220999931 is a deficient number, since it is larger than the sum of its proper divisors (1).

15220999931 is an equidigital number, since it uses as much as digits as its factorization.

15220999931 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 393660, while the sum is 50.

The spelling of 15220999931 in words is "fifteen billion, two hundred twenty million, nine hundred ninety-nine thousand, nine hundred thirty-one".