Base | Representation |
---|---|
bin | 1110100111001110110100… |
… | …11001110111001011010111 |
3 | 11012210000021122001220120221 |
4 | 13103213122121313023113 |
5 | 13202442034011212241 |
6 | 152202144511135211 |
7 | 6524425161642412 |
oct | 723473231671327 |
9 | 135700248056527 |
10 | 32134314554071 |
11 | a26a0a490a334 |
12 | 372ba18b14507 |
13 | 14c13396c2002 |
14 | 7d144833ab79 |
15 | 3aad497c65d1 |
hex | 1d39da6772d7 |
32134314554071 has 2 divisors, whose sum is σ = 32134314554072. Its totient is φ = 32134314554070.
The previous prime is 32134314554059. The next prime is 32134314554267. The reversal of 32134314554071 is 17045541343123.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-32134314554071 is a prime.
It is a super-2 number, since 2×321343145540712 (a number of 28 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (32134314554471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16067157277035 + 16067157277036.
It is an arithmetic number, because the mean of its divisors is an integer number (16067157277036).
Almost surely, 232134314554071 is an apocalyptic number.
32134314554071 is a deficient number, since it is larger than the sum of its proper divisors (1).
32134314554071 is an equidigital number, since it uses as much as digits as its factorization.
32134314554071 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 604800, while the sum is 43.
Adding to 32134314554071 its reverse (17045541343123), we get a palindrome (49179855897194).
The spelling of 32134314554071 in words is "thirty-two trillion, one hundred thirty-four billion, three hundred fourteen million, five hundred fifty-four thousand, seventy-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.131 sec. • engine limits •