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54412254017053 is a prime number
BaseRepresentation
bin11000101111100110101110…
…00100100111011000011101
321010122202112112111220022021
430113303113010213120131
524112442244012021203
6311420351105420141
714314104243644422
oct1427632704473035
9233582475456267
1054412254017053
111637911a960479
12612956193a651
13244909080c4ba
14d617d236bd49
156455c320d5bd
hex317cd712761d

54412254017053 has 2 divisors, whose sum is σ = 54412254017054. Its totient is φ = 54412254017052.

The previous prime is 54412254017029. The next prime is 54412254017063. The reversal of 54412254017053 is 35071045221445.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 32470543113849 + 21941710903204 = 5698293^2 + 4684198^2 .

It is a cyclic number.

It is not a de Polignac number, because 54412254017053 - 29 = 54412254016541 is a prime.

It is a super-3 number, since 3×544122540170533 (a number of 42 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (54412254017063) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27206127008526 + 27206127008527.

It is an arithmetic number, because the mean of its divisors is an integer number (27206127008527).

Almost surely, 254412254017053 is an apocalyptic number.

It is an amenable number.

54412254017053 is a deficient number, since it is larger than the sum of its proper divisors (1).

54412254017053 is an equidigital number, since it uses as much as digits as its factorization.

54412254017053 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 672000, while the sum is 43.

Adding to 54412254017053 its reverse (35071045221445), we get a palindrome (89483299238498).

The spelling of 54412254017053 in words is "fifty-four trillion, four hundred twelve billion, two hundred fifty-four million, seventeen thousand, fifty-three".