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130100119760857 is a prime number
BaseRepresentation
bin11101100101001101001011…
…011011000110111111011001
3122001122110011001000010000211
4131211031023123012333121
5114023030021124321412
61140411100411531121
736255265151504131
oct3545151333067731
9561573131003024
10130100119760857
11384aa1729a1444
1212712336a59aa1
135779508b36a2c
14241ac450b5ac1
15100930908cca7
hex76534b6c6fd9

130100119760857 has 2 divisors, whose sum is σ = 130100119760858. Its totient is φ = 130100119760856.

The previous prime is 130100119760833. The next prime is 130100119760861. The reversal of 130100119760857 is 758067911001031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 82779808952281 + 47320310808576 = 9098341^2 + 6878976^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-130100119760857 is a prime.

It is a super-3 number, since 3×1301001197608573 (a number of 43 digits) contains 333 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 130100119760857.

It is not a weakly prime, because it can be changed into another prime (130100119760357) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65050059880428 + 65050059880429.

It is an arithmetic number, because the mean of its divisors is an integer number (65050059880429).

Almost surely, 2130100119760857 is an apocalyptic number.

It is an amenable number.

130100119760857 is a deficient number, since it is larger than the sum of its proper divisors (1).

130100119760857 is an equidigital number, since it uses as much as digits as its factorization.

130100119760857 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 317520, while the sum is 49.

The spelling of 130100119760857 in words is "one hundred thirty trillion, one hundred billion, one hundred nineteen million, seven hundred sixty thousand, eight hundred fifty-seven".