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485035684072273 is a prime number
BaseRepresentation
bin110111001001000110010101…
…1101001110000101101010001
32100121100221212210002221212101
41232102030223221300231101
51002033311040110303043
64435330152301400401
7204110446261324225
oct15622145351605521
92317327783087771
10485035684072273
11130602529893a14
12464971b9269101
1317a8489031b599
1487abbd0c80785
153b11d57de224d
hex1b9232ba70b51

485035684072273 has 2 divisors, whose sum is σ = 485035684072274. Its totient is φ = 485035684072272.

The previous prime is 485035684072267. The next prime is 485035684072343. The reversal of 485035684072273 is 372270486530584.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 437763225075264 + 47272458997009 = 20922792^2 + 6875497^2 .

It is a cyclic number.

It is not a de Polignac number, because 485035684072273 - 237 = 484898245118801 is a prime.

It is a super-3 number, since 3×4850356840722733 (a number of 45 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (485035684072223) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 242517842036136 + 242517842036137.

It is an arithmetic number, because the mean of its divisors is an integer number (242517842036137).

Almost surely, 2485035684072273 is an apocalyptic number.

It is an amenable number.

485035684072273 is a deficient number, since it is larger than the sum of its proper divisors (1).

485035684072273 is an equidigital number, since it uses as much as digits as its factorization.

485035684072273 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 270950400, while the sum is 64.

The spelling of 485035684072273 in words is "four hundred eighty-five trillion, thirty-five billion, six hundred eighty-four million, seventy-two thousand, two hundred seventy-three".