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13405504331419 is a prime number
BaseRepresentation
bin1100001100010011011001…
…0100101101111010011011
31202110112221122102220110221
43003010312110231322123
53224113433102101134
644302223105420511
72552341615504024
oct303046624557233
952415848386427
1013405504331419
1142a9273794822
1216060b033a737
13763197795aa5
14344b8924bc4b
15183a944085b4
hexc313652de9b

13405504331419 has 2 divisors, whose sum is σ = 13405504331420. Its totient is φ = 13405504331418.

The previous prime is 13405504331413. The next prime is 13405504331437. The reversal of 13405504331419 is 91413340550431.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 13405504331419 - 213 = 13405504323227 is a prime.

It is a super-2 number, since 2×134055043314192 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (13405504331413) 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, 6702752165709 + 6702752165710.

It is an arithmetic number, because the mean of its divisors is an integer number (6702752165710).

Almost surely, 213405504331419 is an apocalyptic number.

13405504331419 is a deficient number, since it is larger than the sum of its proper divisors (1).

13405504331419 is an equidigital number, since it uses as much as digits as its factorization.

13405504331419 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 388800, while the sum is 43.

The spelling of 13405504331419 in words is "thirteen trillion, four hundred five billion, five hundred four million, three hundred thirty-one thousand, four hundred nineteen".