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52341154000153 is a prime number
BaseRepresentation
bin10111110011010100111111…
…11001000010110100011001
320212022202200000102221221221
423321222133321002310121
523330024140411001103
6303153105540332041
714011344406163344
oct1371523771026431
9225282600387857
1052341154000153
111574a8389a1647
125a540966a4021
132328998c75215
14ccd47ab6145b
1560b7a82841bd
hex2f9a9fe42d19

52341154000153 has 2 divisors, whose sum is σ = 52341154000154. Its totient is φ = 52341154000152.

The previous prime is 52341154000151. The next prime is 52341154000199. The reversal of 52341154000153 is 35100045114325.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 35914766509449 + 16426387490704 = 5992893^2 + 4052948^2 .

It is a cyclic number.

It is not a de Polignac number, because 52341154000153 - 21 = 52341154000151 is a prime.

Together with 52341154000151, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (52341154000151) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26170577000076 + 26170577000077.

It is an arithmetic number, because the mean of its divisors is an integer number (26170577000077).

Almost surely, 252341154000153 is an apocalyptic number.

It is an amenable number.

52341154000153 is a deficient number, since it is larger than the sum of its proper divisors (1).

52341154000153 is an equidigital number, since it uses as much as digits as its factorization.

52341154000153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 36000, while the sum is 34.

Adding to 52341154000153 its reverse (35100045114325), we get a palindrome (87441199114478).

The spelling of 52341154000153 in words is "fifty-two trillion, three hundred forty-one billion, one hundred fifty-four million, one hundred fifty-three", and thus it is an aban number.