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50343133994761 is a prime number
BaseRepresentation
bin10110111001001011011001…
…00111100101111100001001
320121020202102120022020222111
423130211230213211330021
523044310214300313021
6255023152205422321
713414112535306316
oct1334455447457411
9217222376266874
1050343133994761
111504a448774043
125790a0586a9a1
13221244c9a24ba
14c6089923dd0d
155c48190168e1
hex2dc96c9e5f09

50343133994761 has 2 divisors, whose sum is σ = 50343133994762. Its totient is φ = 50343133994760.

The previous prime is 50343133994707. The next prime is 50343133994767. The reversal of 50343133994761 is 16749933134305.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 27595385303161 + 22747748691600 = 5253131^2 + 4769460^2 .

It is a cyclic number.

It is not a de Polignac number, because 50343133994761 - 223 = 50343125606153 is a prime.

It is a super-2 number, since 2×503431339947612 (a number of 28 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 50343133994696 and 50343133994705.

It is not a weakly prime, because it can be changed into another prime (50343133994767) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25171566997380 + 25171566997381.

It is an arithmetic number, because the mean of its divisors is an integer number (25171566997381).

Almost surely, 250343133994761 is an apocalyptic number.

It is an amenable number.

50343133994761 is a deficient number, since it is larger than the sum of its proper divisors (1).

50343133994761 is an equidigital number, since it uses as much as digits as its factorization.

50343133994761 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 22044960, while the sum is 58.

The spelling of 50343133994761 in words is "fifty trillion, three hundred forty-three billion, one hundred thirty-three million, nine hundred ninety-four thousand, seven hundred sixty-one".