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100255403767489 is a prime number
BaseRepresentation
bin10110110010111010000111…
…101110100111011011000001
3111010222022201110120012211121
4112302322013232213123001
5101120041031331024424
6553120410251512241
730055133236302205
oct2662720756473301
9433868643505747
10100255403767489
1129a43094856973
12b2b2203a58681
1343c30758c3a99
141aa855d18ab05
15b8cd16be9ee4
hex5b2e87ba76c1

100255403767489 has 2 divisors, whose sum is σ = 100255403767490. Its totient is φ = 100255403767488.

The previous prime is 100255403767447. The next prime is 100255403767609. The reversal of 100255403767489 is 984767304552001.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 99755809255089 + 499594512400 = 9987783^2 + 706820^2 .

It is a cyclic number.

It is not a de Polignac number, because 100255403767489 - 223 = 100255395378881 is a prime.

It is a super-2 number, since 2×1002554037674892 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (100255403777489) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50127701883744 + 50127701883745.

It is an arithmetic number, because the mean of its divisors is an integer number (50127701883745).

It is a 1-persistent number, because it is pandigital, but 2⋅100255403767489 = 200510807534978 is not.

Almost surely, 2100255403767489 is an apocalyptic number.

It is an amenable number.

100255403767489 is a deficient number, since it is larger than the sum of its proper divisors (1).

100255403767489 is an equidigital number, since it uses as much as digits as its factorization.

100255403767489 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 50803200, while the sum is 61.

The spelling of 100255403767489 in words is "one hundred trillion, two hundred fifty-five billion, four hundred three million, seven hundred sixty-seven thousand, four hundred eighty-nine".