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4434312440213 is a prime number
BaseRepresentation
bin100000010000111000110…
…0111100010100110010101
3120200220201212011122010222
41000201301213202212111
51040122432441041323
613233032304524125
7635240252561015
oct100416147424625
916626655148128
104434312440213
11145a643852177
125b7496320645
13262200085826
1411489c591c45
157a52edbc9c8
hex408719e2995

4434312440213 has 2 divisors, whose sum is σ = 4434312440214. Its totient is φ = 4434312440212.

The previous prime is 4434312440209. The next prime is 4434312440231. The reversal of 4434312440213 is 3120442134344.

Together with next prime (4434312440231) it forms an Ormiston pair, because they use the same digits, order apart.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 4282884057169 + 151428383044 = 2069513^2 + 389138^2 .

It is a cyclic number.

It is not a de Polignac number, because 4434312440213 - 22 = 4434312440209 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (4434312440713) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2217156220106 + 2217156220107.

It is an arithmetic number, because the mean of its divisors is an integer number (2217156220107).

Almost surely, 24434312440213 is an apocalyptic number.

It is an amenable number.

4434312440213 is a deficient number, since it is larger than the sum of its proper divisors (1).

4434312440213 is an equidigital number, since it uses as much as digits as its factorization.

4434312440213 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 110592, while the sum is 35.

Adding to 4434312440213 its reverse (3120442134344), we get a palindrome (7554754574557).

The spelling of 4434312440213 in words is "four trillion, four hundred thirty-four billion, three hundred twelve million, four hundred forty thousand, two hundred thirteen".