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54336534041761 is a prime number
BaseRepresentation
bin11000101101011001101011…
…10011110101010010100001
321010101112002111000121102221
430112230311303311102201
524110222210203314021
6311321505323511041
714305453652532106
oct1426546563652241
9233345074017387
1054336534041761
1116349aa3a092a4
12611694b388481
132441bb4321aaa
14d5bc8bbdaaad
1564364087a241
hex316b35cf54a1

54336534041761 has 2 divisors, whose sum is σ = 54336534041762. Its totient is φ = 54336534041760.

The previous prime is 54336534041651. The next prime is 54336534041771. The reversal of 54336534041761 is 16714043563345.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 54312198041761 + 24336000000 = 7369681^2 + 156000^2 .

It is a cyclic number.

It is not a de Polignac number, because 54336534041761 - 217 = 54336533910689 is a prime.

It is a super-2 number, since 2×543365340417612 (a number of 28 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 54336534041699 and 54336534041708.

It is not a weakly prime, because it can be changed into another prime (54336534041771) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27168267020880 + 27168267020881.

It is an arithmetic number, because the mean of its divisors is an integer number (27168267020881).

Almost surely, 254336534041761 is an apocalyptic number.

It is an amenable number.

54336534041761 is a deficient number, since it is larger than the sum of its proper divisors (1).

54336534041761 is an equidigital number, since it uses as much as digits as its factorization.

54336534041761 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 10886400, while the sum is 52.

The spelling of 54336534041761 in words is "fifty-four trillion, three hundred thirty-six billion, five hundred thirty-four million, forty-one thousand, seven hundred sixty-one".