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340276948125019 is a prime number
BaseRepresentation
bin100110101011110101110011…
…0100111100100010101011011
31122121211011021120122001020021
41031113223212213210111123
5324100044142210000034
63203413023312011311
7131450124663002533
oct11527534647442533
91577734246561207
10340276948125019
119947170923aa44
12321b7b51530b37
13117b3c94a78369
1460056c67004c3
1529515b70ce6b4
hex1357ae69e455b

340276948125019 has 2 divisors, whose sum is σ = 340276948125020. Its totient is φ = 340276948125018.

The previous prime is 340276948124987. The next prime is 340276948125029. The reversal of 340276948125019 is 910521849672043.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 340276948125019 - 25 = 340276948124987 is a prime.

It is a super-3 number, since 3×3402769481250193 (a number of 45 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (340276948125029) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 170138474062509 + 170138474062510.

It is an arithmetic number, because the mean of its divisors is an integer number (170138474062510).

It is a 1-persistent number, because it is pandigital, but 2⋅340276948125019 = 680553896250038 is not.

Almost surely, 2340276948125019 is an apocalyptic number.

340276948125019 is a deficient number, since it is larger than the sum of its proper divisors (1).

340276948125019 is an equidigital number, since it uses as much as digits as its factorization.

340276948125019 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 26127360, while the sum is 61.

The spelling of 340276948125019 in words is "three hundred forty trillion, two hundred seventy-six billion, nine hundred forty-eight million, one hundred twenty-five thousand, nineteen".