Base | Representation |
---|---|
bin | 111111001010111010010… |
… | …101011010000111011011 |
3 | 120100222222202112102112112 |
4 | 333022322111122013123 |
5 | 1032110424021040011 |
6 | 13122125300102535 |
7 | 625426044616052 |
oct | 77127225320733 |
9 | 16328882472475 |
10 | 4341043143131 |
11 | 1424031757872 |
12 | 5a13a6694a4b |
13 | 256487b41146 |
14 | 1101713cb399 |
15 | 77dc1a0558b |
hex | 3f2ba55a1db |
4341043143131 has 2 divisors, whose sum is σ = 4341043143132. Its totient is φ = 4341043143130.
The previous prime is 4341043143121. The next prime is 4341043143139. The reversal of 4341043143131 is 1313413401434.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 4341043143131 - 218 = 4341042880987 is a prime.
It is a super-2 number, since 2×43410431431312 (a number of 26 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (4341043143139) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2170521571565 + 2170521571566.
It is an arithmetic number, because the mean of its divisors is an integer number (2170521571566).
Almost surely, 24341043143131 is an apocalyptic number.
4341043143131 is a deficient number, since it is larger than the sum of its proper divisors (1).
4341043143131 is an equidigital number, since it uses as much as digits as its factorization.
4341043143131 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 20736, while the sum is 32.
Adding to 4341043143131 its reverse (1313413401434), we get a palindrome (5654456544565).
The spelling of 4341043143131 in words is "four trillion, three hundred forty-one billion, forty-three million, one hundred forty-three thousand, one hundred thirty-one".
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