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2541320551139 is a prime number
BaseRepresentation
bin100100111110110010100…
…000100011101011100011
322222221120222222001022112
4210332302200203223203
5313114111030114024
65223244435025535
7351414202161506
oct44766240435343
98887528861275
102541320551139
1189a84a2209a7
1235063713b2ab
131558512168b6
148b0017d113d
154618b4e700e
hex24fb2823ae3

2541320551139 has 2 divisors, whose sum is σ = 2541320551140. Its totient is φ = 2541320551138.

The previous prime is 2541320551067. The next prime is 2541320551163. The reversal of 2541320551139 is 9311550231452.

2541320551139 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 2541320551139 - 216 = 2541320485603 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 2541320551096 and 2541320551105.

It is not a weakly prime, because it can be changed into another prime (2541320511139) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1270660275569 + 1270660275570.

It is an arithmetic number, because the mean of its divisors is an integer number (1270660275570).

Almost surely, 22541320551139 is an apocalyptic number.

2541320551139 is a deficient number, since it is larger than the sum of its proper divisors (1).

2541320551139 is an equidigital number, since it uses as much as digits as its factorization.

2541320551139 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 162000, while the sum is 41.

The spelling of 2541320551139 in words is "two trillion, five hundred forty-one billion, three hundred twenty million, five hundred fifty-one thousand, one hundred thirty-nine".