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14203113199883 is a prime number
BaseRepresentation
bin1100111010101110101110…
…0001000101100100001011
31212021210201101020120122022
43032223223201011210023
53330200433434344013
650112453051515055
72664066246150056
oct316535341054413
955253641216568
1014203113199883
114586563143a53
1217147a9a58a8b
137c046a24031a
14371611bc1a9d
151996c7827008
hexceaeb84590b

14203113199883 has 2 divisors, whose sum is σ = 14203113199884. Its totient is φ = 14203113199882.

The previous prime is 14203113199789. The next prime is 14203113199903. The reversal of 14203113199883 is 38899131130241.

14203113199883 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 14203113199883 - 226 = 14203046091019 is a prime.

It is a super-2 number, since 2×142031131998832 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (14203113195883) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7101556599941 + 7101556599942.

It is an arithmetic number, because the mean of its divisors is an integer number (7101556599942).

Almost surely, 214203113199883 is an apocalyptic number.

14203113199883 is a deficient number, since it is larger than the sum of its proper divisors (1).

14203113199883 is an equidigital number, since it uses as much as digits as its factorization.

14203113199883 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1119744, while the sum is 53.

The spelling of 14203113199883 in words is "fourteen trillion, two hundred three billion, one hundred thirteen million, one hundred ninety-nine thousand, eight hundred eighty-three".