Base | Representation |
---|---|
bin | 11111010110000111101… |
… | …000010101100010011001 |
3 | 21121220222201211122222000 |
4 | 133112013220111202121 |
5 | 240243000323112213 |
6 | 4325320411151213 |
7 | 311424316602015 |
oct | 37260750254231 |
9 | 7556881748860 |
10 | 2154054113433 |
11 | 760590433402 |
12 | 2a9578445509 |
13 | 128184a00a01 |
14 | 7638483d545 |
15 | 3b072970c73 |
hex | 1f587a15899 |
2154054113433 has 32 divisors (see below), whose sum is σ = 3331779194880. Its totient is φ = 1372840531920.
The previous prime is 2154054113383. The next prime is 2154054113441. The reversal of 2154054113433 is 3343114504512.
2154054113433 is a `hidden beast` number, since 2 + 1 + 54 + 0 + 541 + 1 + 34 + 33 = 666.
It is not a de Polignac number, because 2154054113433 - 29 = 2154054112921 is a prime.
It is not an unprimeable number, because it can be changed into a prime (2154054113443) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 166960 + ... + 2082302.
It is an arithmetic number, because the mean of its divisors is an integer number (104118099840).
Almost surely, 22154054113433 is an apocalyptic number.
2154054113433 is a gapful number since it is divisible by the number (23) formed by its first and last digit.
It is an amenable number.
2154054113433 is a deficient number, since it is larger than the sum of its proper divisors (1177725081447).
2154054113433 is a wasteful number, since it uses less digits than its factorization.
2154054113433 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1917186 (or 1917180 counting only the distinct ones).
The product of its (nonzero) digits is 86400, while the sum is 36.
Adding to 2154054113433 its reverse (3343114504512), we get a palindrome (5497168617945).
The spelling of 2154054113433 in words is "two trillion, one hundred fifty-four billion, fifty-four million, one hundred thirteen thousand, four hundred thirty-three".
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