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13311430512143 is a prime number
BaseRepresentation
bin1100000110110100111100…
…0101100100111000001111
31202010120010110221201220012
43001231033011210320033
53221043302202342033
644151104201143435
72542501441312241
oct301551705447017
952116113851805
1013311430512143
11427238954aa45
1215aba1718ab7b
1375735497707a
143403c32a2091
151813da5c4248
hexc1b4f164e0f

13311430512143 has 2 divisors, whose sum is σ = 13311430512144. Its totient is φ = 13311430512142.

The previous prime is 13311430512139. The next prime is 13311430512167. The reversal of 13311430512143 is 34121503411331.

13311430512143 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 13311430512143 - 22 = 13311430512139 is a prime.

It is a super-2 number, since 2×133114305121432 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13311430532143) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6655715256071 + 6655715256072.

It is an arithmetic number, because the mean of its divisors is an integer number (6655715256072).

Almost surely, 213311430512143 is an apocalyptic number.

13311430512143 is a deficient number, since it is larger than the sum of its proper divisors (1).

13311430512143 is an equidigital number, since it uses as much as digits as its factorization.

13311430512143 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 12960, while the sum is 32.

Adding to 13311430512143 its reverse (34121503411331), we get a palindrome (47432933923474).

The spelling of 13311430512143 in words is "thirteen trillion, three hundred eleven billion, four hundred thirty million, five hundred twelve thousand, one hundred forty-three".