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15100410412121 is a prime number
BaseRepresentation
bin1101101110111101011010…
…0110010011110001011001
31222110120210120020211002012
43123233112212103301121
53434401120031141441
652041011025115305
73115653150463646
oct333572646236131
958416716224065
1015100410412121
1148a2059013351
12183a688a88b35
13856c68a77b18
143a2c1424adcd
151b2be291c9eb
hexdbbd6993c59

15100410412121 has 2 divisors, whose sum is σ = 15100410412122. Its totient is φ = 15100410412120.

The previous prime is 15100410412039. The next prime is 15100410412177. The reversal of 15100410412121 is 12121401400151.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 14594233657600 + 506176754521 = 3820240^2 + 711461^2 .

It is a cyclic number.

It is not a de Polignac number, because 15100410412121 - 218 = 15100410149977 is a prime.

It is a super-2 number, since 2×151004104121212 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (15100410412721) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7550205206060 + 7550205206061.

It is an arithmetic number, because the mean of its divisors is an integer number (7550205206061).

Almost surely, 215100410412121 is an apocalyptic number.

It is an amenable number.

15100410412121 is a deficient number, since it is larger than the sum of its proper divisors (1).

15100410412121 is an equidigital number, since it uses as much as digits as its factorization.

15100410412121 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 320, while the sum is 23.

Adding to 15100410412121 its reverse (12121401400151), we get a palindrome (27221811812272).

The spelling of 15100410412121 in words is "fifteen trillion, one hundred billion, four hundred ten million, four hundred twelve thousand, one hundred twenty-one".