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155511504143 is a prime number
BaseRepresentation
bin1001000011010100110…
…1010011110100001111
3112212101212022221012122
42100311031103310033
510021441421113033
6155235131123155
714143501655312
oct2206515236417
9485355287178
10155511504143
115aa5220a856
12261805a34bb
1311884390508
147753760979
1540a28c2a98
hex2435353d0f

155511504143 has 2 divisors, whose sum is σ = 155511504144. Its totient is φ = 155511504142.

The previous prime is 155511504139. The next prime is 155511504173. The reversal of 155511504143 is 341405115551.

155511504143 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is an emirp because it is prime and its reverse (341405115551) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 155511504143 - 22 = 155511504139 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (155511504173) 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, 77755752071 + 77755752072.

It is an arithmetic number, because the mean of its divisors is an integer number (77755752072).

Almost surely, 2155511504143 is an apocalyptic number.

155511504143 is a deficient number, since it is larger than the sum of its proper divisors (1).

155511504143 is an equidigital number, since it uses as much as digits as its factorization.

155511504143 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 30000, while the sum is 35.

Adding to 155511504143 its reverse (341405115551), we get a palindrome (496916619694).

The spelling of 155511504143 in words is "one hundred fifty-five billion, five hundred eleven million, five hundred four thousand, one hundred forty-three".