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55146345254353 is a prime number
BaseRepresentation
bin11001000100111110000100…
…10100011100110111010001
321020020221100121200120220201
430202133002110130313101
524212004203341114403
6313141514204335201
714421122562341353
oct1442370224346721
9236227317616821
1055146345254353
1116631485263251
126227893544501
1324a037cb290b2
14d891511c10d3
1565973a18821d
hex3227c251cdd1

55146345254353 has 2 divisors, whose sum is σ = 55146345254354. Its totient is φ = 55146345254352.

The previous prime is 55146345254303. The next prime is 55146345254401. The reversal of 55146345254353 is 35345254364155.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 54177593254849 + 968751999504 = 7360543^2 + 984252^2 .

It is a cyclic number.

It is not a de Polignac number, because 55146345254353 - 29 = 55146345253841 is a prime.

It is a super-2 number, since 2×551463452543532 (a number of 28 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 55146345254294 and 55146345254303.

It is not a weakly prime, because it can be changed into another prime (55146345254303) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27573172627176 + 27573172627177.

It is an arithmetic number, because the mean of its divisors is an integer number (27573172627177).

Almost surely, 255146345254353 is an apocalyptic number.

It is an amenable number.

55146345254353 is a deficient number, since it is larger than the sum of its proper divisors (1).

55146345254353 is an equidigital number, since it uses as much as digits as its factorization.

55146345254353 is an evil number, because the sum of its binary digits is even.

The product of its digits is 64800000, while the sum is 55.

The spelling of 55146345254353 in words is "fifty-five trillion, one hundred forty-six billion, three hundred forty-five million, two hundred fifty-four thousand, three hundred fifty-three".