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15100506000553 is a prime number
BaseRepresentation
bin1101101110111101110001…
…0010111100110010101001
31222110121001020010012202111
43123233130102330302221
53434401314014004203
652041024322002321
73115655421123301
oct333573422746251
958417036105674
1015100506000553
1148a20a7a71356
12183a6b4aa63a1
13856c8280561c
143a2c22c10601
151b2beb00026d
hexdbbdc4bcca9

15100506000553 has 2 divisors, whose sum is σ = 15100506000554. Its totient is φ = 15100506000552.

The previous prime is 15100506000551. The next prime is 15100506000571. The reversal of 15100506000553 is 35500060500151.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 12857301861264 + 2243204139289 = 3585708^2 + 1497733^2 .

It is a cyclic number.

It is not a de Polignac number, because 15100506000553 - 21 = 15100506000551 is a prime.

It is a super-3 number, since 3×151005060005533 (a number of 41 digits) contains 333 as substring.

Together with 15100506000551, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (15100506000551) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7550253000276 + 7550253000277.

It is an arithmetic number, because the mean of its divisors is an integer number (7550253000277).

Almost surely, 215100506000553 is an apocalyptic number.

It is an amenable number.

15100506000553 is a deficient number, since it is larger than the sum of its proper divisors (1).

15100506000553 is an equidigital number, since it uses as much as digits as its factorization.

15100506000553 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 11250, while the sum is 31.

The spelling of 15100506000553 in words is "fifteen trillion, one hundred billion, five hundred six million, five hundred fifty-three", and thus it is an aban number.