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311431244421433 is a prime number
BaseRepresentation
bin100011011001111101011110…
…0101100100001100100111001
31111211200112110220001200211011
41012303322330230201210321
5311304442142032441213
63022205312200233521
7122412110363033203
oct10663727454414471
91454615426050734
10311431244421433
1190260281531024
122ab1956576b8a1
13104a0ac02a7926
1456c94d198d773
1526010948ed03d
hex11b3ebcb21939

311431244421433 has 2 divisors, whose sum is σ = 311431244421434. Its totient is φ = 311431244421432.

The previous prime is 311431244421391. The next prime is 311431244421457. The reversal of 311431244421433 is 334124442134113.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 289954248409849 + 21476996011584 = 17028043^2 + 4634328^2 .

It is a cyclic number.

It is not a de Polignac number, because 311431244421433 - 29 = 311431244420921 is a prime.

It is not a weakly prime, because it can be changed into another prime (311431244421733) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155715622210716 + 155715622210717.

It is an arithmetic number, because the mean of its divisors is an integer number (155715622210717).

Almost surely, 2311431244421433 is an apocalyptic number.

It is an amenable number.

311431244421433 is a deficient number, since it is larger than the sum of its proper divisors (1).

311431244421433 is an equidigital number, since it uses as much as digits as its factorization.

311431244421433 is an evil number, because the sum of its binary digits is even.

The product of its digits is 331776, while the sum is 40.

Adding to 311431244421433 its reverse (334124442134113), we get a palindrome (645555686555546).

The spelling of 311431244421433 in words is "three hundred eleven trillion, four hundred thirty-one billion, two hundred forty-four million, four hundred twenty-one thousand, four hundred thirty-three".