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4483326171253 is a prime number
BaseRepresentation
bin100000100111101101100…
…0100000101100001110101
3120212121020201011011002211
41001033123010011201311
51041423322444440003
613311340035133421
7641624000452036
oct101173304054165
916777221134084
104483326171253
11147940581a806
12604a94a44271
13266a116b5247
14116dcbd0aa8d
157b94cd9496d
hex413db105875

4483326171253 has 2 divisors, whose sum is σ = 4483326171254. Its totient is φ = 4483326171252.

The previous prime is 4483326171223. The next prime is 4483326171257. The reversal of 4483326171253 is 3521716233844.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 4068619794724 + 414706376529 = 2017082^2 + 643977^2 .

It is a cyclic number.

It is not a de Polignac number, because 4483326171253 - 25 = 4483326171221 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 4483326171197 and 4483326171206.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (4483326171257) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2241663085626 + 2241663085627.

It is an arithmetic number, because the mean of its divisors is an integer number (2241663085627).

Almost surely, 24483326171253 is an apocalyptic number.

It is an amenable number.

4483326171253 is a deficient number, since it is larger than the sum of its proper divisors (1).

4483326171253 is an equidigital number, since it uses as much as digits as its factorization.

4483326171253 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 2903040, while the sum is 49.

The spelling of 4483326171253 in words is "four trillion, four hundred eighty-three billion, three hundred twenty-six million, one hundred seventy-one thousand, two hundred fifty-three".