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6331434341341 is a prime number
BaseRepresentation
bin101110000100010011011…
…1001010001001111011101
3211102021112112201120102211
41130020212321101033131
51312213234142410331
621244342031120421
71222300562634544
oct134104671211735
924367475646384
106331434341341
11202116769764a
128630a8673111
1336c089767285
1417c62ba6b15b
15aea6614a8b1
hex5c226e513dd

6331434341341 has 2 divisors, whose sum is σ = 6331434341342. Its totient is φ = 6331434341340.

The previous prime is 6331434341311. The next prime is 6331434341357. The reversal of 6331434341341 is 1431434341336.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 3858533419225 + 2472900922116 = 1964315^2 + 1572546^2 .

It is a cyclic number.

It is not a de Polignac number, because 6331434341341 - 213 = 6331434333149 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 6331434341294 and 6331434341303.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (6331434341311) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3165717170670 + 3165717170671.

It is an arithmetic number, because the mean of its divisors is an integer number (3165717170671).

Almost surely, 26331434341341 is an apocalyptic number.

It is an amenable number.

6331434341341 is a deficient number, since it is larger than the sum of its proper divisors (1).

6331434341341 is an equidigital number, since it uses as much as digits as its factorization.

6331434341341 is an evil number, because the sum of its binary digits is even.

The product of its digits is 373248, while the sum is 40.

Adding to 6331434341341 its reverse (1431434341336), we get a palindrome (7762868682677).

The spelling of 6331434341341 in words is "six trillion, three hundred thirty-one billion, four hundred thirty-four million, three hundred forty-one thousand, three hundred forty-one".