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3161450340233 is a prime number
BaseRepresentation
bin101110000000010101000…
…111111101111110001001
3102012020020220120100012222
4232000111013331332021
5403244122241341413
610420203352343425
7444256454335343
oct56002507757611
912166226510188
103161450340233
111009844330a29
124308633a4b75
1319c17b3c24c3
14ad02d2a4293
15573836bbb08
hex2e0151fdf89

3161450340233 has 2 divisors, whose sum is σ = 3161450340234. Its totient is φ = 3161450340232.

The previous prime is 3161450340227. The next prime is 3161450340263. The reversal of 3161450340233 is 3320430541613.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 3114349621504 + 47100718729 = 1764752^2 + 217027^2 .

It is an emirp because it is prime and its reverse (3320430541613) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 3161450340233 - 212 = 3161450336137 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 3161450340193 and 3161450340202.

It is not a weakly prime, because it can be changed into another prime (3161450340263) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1580725170116 + 1580725170117.

It is an arithmetic number, because the mean of its divisors is an integer number (1580725170117).

Almost surely, 23161450340233 is an apocalyptic number.

It is an amenable number.

3161450340233 is a deficient number, since it is larger than the sum of its proper divisors (1).

3161450340233 is an equidigital number, since it uses as much as digits as its factorization.

3161450340233 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 77760, while the sum is 35.

Adding to 3161450340233 its reverse (3320430541613), we get a palindrome (6481880881846).

The spelling of 3161450340233 in words is "three trillion, one hundred sixty-one billion, four hundred fifty million, three hundred forty thousand, two hundred thirty-three".