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139542519241 is a prime number
BaseRepresentation
bin1000000111110101100…
…0011101110111001001
3111100011221212110010101
42001331120131313021
54241240341103431
6144034400245401
713036664562106
oct2017530356711
9440157773111
10139542519241
11541a8129074
1223064610861
131020ab76318
146a7a983aad
15396a9a6961
hex207d61ddc9

139542519241 has 2 divisors, whose sum is σ = 139542519242. Its totient is φ = 139542519240.

The previous prime is 139542519227. The next prime is 139542519247. The reversal of 139542519241 is 142915245931.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 117703886400 + 21838632841 = 343080^2 + 147779^2 .

It is a cyclic number.

It is not a de Polignac number, because 139542519241 - 25 = 139542519209 is a prime.

It is a super-3 number, since 3×1395425192413 (a number of 34 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 139542519191 and 139542519200.

It is not a weakly prime, because it can be changed into another prime (139542519247) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 69771259620 + 69771259621.

It is an arithmetic number, because the mean of its divisors is an integer number (69771259621).

Almost surely, 2139542519241 is an apocalyptic number.

It is an amenable number.

139542519241 is a deficient number, since it is larger than the sum of its proper divisors (1).

139542519241 is an equidigital number, since it uses as much as digits as its factorization.

139542519241 is an evil number, because the sum of its binary digits is even.

The product of its digits is 388800, while the sum is 46.

The spelling of 139542519241 in words is "one hundred thirty-nine billion, five hundred forty-two million, five hundred nineteen thousand, two hundred forty-one".