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64153419212903 is a prime number
BaseRepresentation
bin11101001011000111000011…
…10000110110100001100111
322102011000012002101202002212
432211203201300312201213
531402042200304303103
6344235401322354035
716340634350640536
oct1645434160664147
9272130162352085
1064153419212903
1119494348740109
12724144235b91b
1329a4845418277
1411bb08359631d
15763b9e8dc7d8
hex3a58e1c36867

64153419212903 has 2 divisors, whose sum is σ = 64153419212904. Its totient is φ = 64153419212902.

The previous prime is 64153419212821. The next prime is 64153419212941. The reversal of 64153419212903 is 30921291435146.

64153419212903 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 64153419212903 - 234 = 64136239343719 is a prime.

It is a super-2 number, since 2×641534192129032 (a number of 28 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (64153419217903) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32076709606451 + 32076709606452.

It is an arithmetic number, because the mean of its divisors is an integer number (32076709606452).

Almost surely, 264153419212903 is an apocalyptic number.

64153419212903 is a deficient number, since it is larger than the sum of its proper divisors (1).

64153419212903 is an equidigital number, since it uses as much as digits as its factorization.

64153419212903 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1399680, while the sum is 50.

The spelling of 64153419212903 in words is "sixty-four trillion, one hundred fifty-three billion, four hundred nineteen million, two hundred twelve thousand, nine hundred three".