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131132022017 is a prime number
BaseRepresentation
bin111101000100000010…
…0111101110100000001
3110112110211000220220012
41322020010331310001
54122024244201032
6140124034154305
712321400540043
oct1721004756401
9415424026805
10131132022017
1150681676017
12214b7a22995
13c49a587949
1464bd981a93
1536274226b2
hex1e8813dd01

131132022017 has 2 divisors, whose sum is σ = 131132022018. Its totient is φ = 131132022016.

The previous prime is 131132021989. The next prime is 131132022023. The reversal of 131132022017 is 710220231131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 117764962561 + 13367059456 = 343169^2 + 115616^2 .

It is a cyclic number.

It is not a de Polignac number, because 131132022017 - 224 = 131115244801 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 131132021983 and 131132022001.

It is not a weakly prime, because it can be changed into another prime (131132028017) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65566011008 + 65566011009.

It is an arithmetic number, because the mean of its divisors is an integer number (65566011009).

Almost surely, 2131132022017 is an apocalyptic number.

It is an amenable number.

131132022017 is a deficient number, since it is larger than the sum of its proper divisors (1).

131132022017 is an equidigital number, since it uses as much as digits as its factorization.

131132022017 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 504, while the sum is 23.

Adding to 131132022017 its reverse (710220231131), we get a palindrome (841352253148).

The spelling of 131132022017 in words is "one hundred thirty-one billion, one hundred thirty-two million, twenty-two thousand, seventeen".