Base | Representation |
---|---|
bin | 101101110100010001… |
… | …010100110110000111 |
3 | 11200222111201011102220 |
4 | 231310101110312013 |
5 | 1301223001323033 |
6 | 34333334503423 |
7 | 3361051324506 |
oct | 556421246607 |
9 | 150874634386 |
10 | 49195339143 |
11 | 19955539485 |
12 | 964b504b73 |
13 | 4840147542 |
14 | 254991083d |
15 | 142ddebbb3 |
hex | b74454d87 |
49195339143 has 24 divisors (see below), whose sum is σ = 69709016400. Its totient is φ = 30854461440.
The previous prime is 49195339129. The next prime is 49195339153. The reversal of 49195339143 is 34193359194.
It is a happy number.
49195339143 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 49195339143 - 25 = 49195339111 is a prime.
It is a super-3 number, since 3×491953391433 (a number of 33 digits) contains 333 as substring.
It is a Harshad number since it is a multiple of its sum of digits (51).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (49195339153) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 2261583 + ... + 2283231.
It is an arithmetic number, because the mean of its divisors is an integer number (2904542350).
Almost surely, 249195339143 is an apocalyptic number.
49195339143 is a deficient number, since it is larger than the sum of its proper divisors (20513677257).
49195339143 is a wasteful number, since it uses less digits than its factorization.
49195339143 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 24307 (or 24290 counting only the distinct ones).
The product of its digits is 1574640, while the sum is 51.
The spelling of 49195339143 in words is "forty-nine billion, one hundred ninety-five million, three hundred thirty-nine thousand, one hundred forty-three".
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