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43011314207 is a prime number
BaseRepresentation
bin101000000011101011…
…001001001000011111
311010000112101011000022
4220003223021020133
51201041344023312
631431545422355
73051601065341
oct500353111037
9133015334008
1043011314207
1117271846a74
12840449a9bb
134095bc88a4
1421204bb291
1511bb063172
hexa03ac921f

43011314207 has 2 divisors, whose sum is σ = 43011314208. Its totient is φ = 43011314206.

The previous prime is 43011314191. The next prime is 43011314209. The reversal of 43011314207 is 70241311034.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 43011314207 - 24 = 43011314191 is a prime.

Together with 43011314209, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (43011314209) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21505657103 + 21505657104.

It is an arithmetic number, because the mean of its divisors is an integer number (21505657104).

Almost surely, 243011314207 is an apocalyptic number.

43011314207 is a deficient number, since it is larger than the sum of its proper divisors (1).

43011314207 is an equidigital number, since it uses as much as digits as its factorization.

43011314207 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2016, while the sum is 26.

The spelling of 43011314207 in words is "forty-three billion, eleven million, three hundred fourteen thousand, two hundred seven".