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115032224243957 is a prime number
BaseRepresentation
bin11010001001111100000110…
…110111010010000011110101
3120002021222021111021010000012
4122021330012313102003311
5110034141443401301312
61044353024013430005
733141542516033066
oct3211740667220365
9502258244233005
10115032224243957
113371a98a77a129
1210a9a02319b305
134c2565249a1a7
1420598372c056d
15d473bae68322
hex689f06dd20f5

115032224243957 has 2 divisors, whose sum is σ = 115032224243958. Its totient is φ = 115032224243956.

The previous prime is 115032224243917. The next prime is 115032224244011. The reversal of 115032224243957 is 759342422230511.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 67658751544036 + 47373472699921 = 8225494^2 + 6882839^2 .

It is a cyclic number.

It is not a de Polignac number, because 115032224243957 - 210 = 115032224242933 is a prime.

It is a super-3 number, since 3×1150322242439573 (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 (115032224243917) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57516112121978 + 57516112121979.

It is an arithmetic number, because the mean of its divisors is an integer number (57516112121979).

Almost surely, 2115032224243957 is an apocalyptic number.

It is an amenable number.

115032224243957 is a deficient number, since it is larger than the sum of its proper divisors (1).

115032224243957 is an equidigital number, since it uses as much as digits as its factorization.

115032224243957 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3628800, while the sum is 50.

The spelling of 115032224243957 in words is "one hundred fifteen trillion, thirty-two billion, two hundred twenty-four million, two hundred forty-three thousand, nine hundred fifty-seven".