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143134120033 is a prime number
BaseRepresentation
bin1000010101001101110…
…1010101100001100001
3111200110021002200011221
42011103131111201201
54321114313320113
6145431020535041
713224665642332
oct2052335254141
9450407080157
10143134120033
115578052a993
12238a73b1481
1310660ca53a8
146cdb988289
153acae5998d
hex2153755861

143134120033 has 2 divisors, whose sum is σ = 143134120034. Its totient is φ = 143134120032.

The previous prime is 143134120027. The next prime is 143134120039. The reversal of 143134120033 is 330021431341.

It is a balanced prime because it is at equal distance from previous prime (143134120027) and next prime (143134120039).

It can be written as a sum of positive squares in only one way, i.e., 98362522384 + 44771597649 = 313628^2 + 211593^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-143134120033 is a prime.

It is not a weakly prime, because it can be changed into another prime (143134120039) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71567060016 + 71567060017.

It is an arithmetic number, because the mean of its divisors is an integer number (71567060017).

Almost surely, 2143134120033 is an apocalyptic number.

It is an amenable number.

143134120033 is a deficient number, since it is larger than the sum of its proper divisors (1).

143134120033 is an equidigital number, since it uses as much as digits as its factorization.

143134120033 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 143134120033 its reverse (330021431341), we get a palindrome (473155551374).

The spelling of 143134120033 in words is "one hundred forty-three billion, one hundred thirty-four million, one hundred twenty thousand, thirty-three".