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250044343331801 is a prime number
BaseRepresentation
bin111000110110100111111100…
…010000110010101111011001
31012210100000011012120222100112
4320312213330100302233121
5230233211303343104201
62243444441402214105
7103445051632115444
oct7066477420625731
91183300135528315
10250044343331801
1172743235193997
12240643287b4335
13a96a107845b15
1445a62d45a265b
151dd9361b790bb
hexe369fc432bd9

250044343331801 has 2 divisors, whose sum is σ = 250044343331802. Its totient is φ = 250044343331800.

The previous prime is 250044343331749. The next prime is 250044343331807. The reversal of 250044343331801 is 108133343440052.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 211563424848400 + 38480918483401 = 14545220^2 + 6203299^2 .

It is a cyclic number.

It is not a de Polignac number, because 250044343331801 - 210 = 250044343330777 is a prime.

It is not a weakly prime, because it can be changed into another prime (250044343331807) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 125022171665900 + 125022171665901.

It is an arithmetic number, because the mean of its divisors is an integer number (125022171665901).

Almost surely, 2250044343331801 is an apocalyptic number.

It is an amenable number.

250044343331801 is a deficient number, since it is larger than the sum of its proper divisors (1).

250044343331801 is an equidigital number, since it uses as much as digits as its factorization.

250044343331801 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 414720, while the sum is 41.

Adding to 250044343331801 its reverse (108133343440052), we get a palindrome (358177686771853).

The spelling of 250044343331801 in words is "two hundred fifty trillion, forty-four billion, three hundred forty-three million, three hundred thirty-one thousand, eight hundred one".