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143643221255581 is a prime number
BaseRepresentation
bin100000101010010010001011…
…001011010110010110011101
3200211121011021001210010222021
4200222102023023112112131
5122311422304120134311
61225300441310220141
742153610634020525
oct4052221313262635
9624534231703867
10143643221255581
114185082569a553
121413b0281b4651
13621c661421938
142768524794685
151191753281171
hex82a48b2d659d

143643221255581 has 2 divisors, whose sum is σ = 143643221255582. Its totient is φ = 143643221255580.

The previous prime is 143643221255467. The next prime is 143643221255629. The reversal of 143643221255581 is 185552122346341.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 134281512240100 + 9361709015481 = 11587990^2 + 3059691^2 .

It is a cyclic number.

It is not a de Polignac number, because 143643221255581 - 219 = 143643220731293 is a prime.

It is a super-3 number, since 3×1436432212555813 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (143643221255981) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71821610627790 + 71821610627791.

It is an arithmetic number, because the mean of its divisors is an integer number (71821610627791).

Almost surely, 2143643221255581 is an apocalyptic number.

It is an amenable number.

143643221255581 is a deficient number, since it is larger than the sum of its proper divisors (1).

143643221255581 is an equidigital number, since it uses as much as digits as its factorization.

143643221255581 is an evil number, because the sum of its binary digits is even.

The product of its digits is 6912000, while the sum is 52.

The spelling of 143643221255581 in words is "one hundred forty-three trillion, six hundred forty-three billion, two hundred twenty-one million, two hundred fifty-five thousand, five hundred eighty-one".