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311210323301131 is a prime number
BaseRepresentation
bin100011011000010110100110…
…0110001010101001100001011
31111210220102020111222222120021
41012300231030301111030023
5311242332220231114011
63021520014314041311
7122360126633350312
oct10660551461251413
91453812214888507
10311210323301131
11901856142a3561
122aaa278b719237
1310486013801137
1456bc93512d279
1525ea46490d471
hex11b0b4cc5530b

311210323301131 has 2 divisors, whose sum is σ = 311210323301132. Its totient is φ = 311210323301130.

The previous prime is 311210323301081. The next prime is 311210323301177. The reversal of 311210323301131 is 131103323012113.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 311210323301131 - 215 = 311210323268363 is a prime.

It is a super-2 number, since 2×3112103233011312 (a number of 30 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 311210323301096 and 311210323301105.

It is not a weakly prime, because it can be changed into another prime (311210323301231) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155605161650565 + 155605161650566.

It is an arithmetic number, because the mean of its divisors is an integer number (155605161650566).

Almost surely, 2311210323301131 is an apocalyptic number.

311210323301131 is a deficient number, since it is larger than the sum of its proper divisors (1).

311210323301131 is an equidigital number, since it uses as much as digits as its factorization.

311210323301131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 972, while the sum is 25.

Adding to 311210323301131 its reverse (131103323012113), we get a palindrome (442313646313244).

The spelling of 311210323301131 in words is "three hundred eleven trillion, two hundred ten billion, three hundred twenty-three million, three hundred one thousand, one hundred thirty-one".