Base | Representation |
---|---|
bin | 1101000001000101111000… |
… | …1000101001000111001001 |
3 | 1212200020211111010021212212 |
4 | 3100101132020221013021 |
5 | 3333443304022312441 |
6 | 50235010333114505 |
7 | 3005015604504365 |
oct | 320213610510711 |
9 | 55606744107785 |
10 | 14312410354121 |
11 | 461894a5806a0 |
12 | 1731a11328a35 |
13 | 7ca867c42759 |
14 | 376a1d8ca2a5 |
15 | 19c472d88eeb |
hex | d045e2291c9 |
14312410354121 has 16 divisors (see below), whose sum is σ = 15642838411200. Its totient is φ = 12986871052800.
The previous prime is 14312410354069. The next prime is 14312410354147. The reversal of 14312410354121 is 12145301421341.
It is a cyclic number.
It is not a de Polignac number, because 14312410354121 - 238 = 14037532447177 is a prime.
It is not an unprimeable number, because it can be changed into a prime (14312410314121) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 54252866 + ... + 54516036.
It is an arithmetic number, because the mean of its divisors is an integer number (977677400700).
Almost surely, 214312410354121 is an apocalyptic number.
14312410354121 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
It is an amenable number.
14312410354121 is a deficient number, since it is larger than the sum of its proper divisors (1330428057079).
14312410354121 is a wasteful number, since it uses less digits than its factorization.
14312410354121 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 272440.
The product of its (nonzero) digits is 11520, while the sum is 32.
Adding to 14312410354121 its reverse (12145301421341), we get a palindrome (26457711775462).
The spelling of 14312410354121 in words is "fourteen trillion, three hundred twelve billion, four hundred ten million, three hundred fifty-four thousand, one hundred twenty-one".
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