Base | Representation |
---|---|
bin | 111111000001111100111… |
… | …011111010001110001000 |
3 | 120100002011022202022121202 |
4 | 333001330323322032020 |
5 | 1031431223304122440 |
6 | 13113454525323332 |
7 | 624635462666204 |
oct | 77017473721610 |
9 | 16302138668552 |
10 | 4331423114120 |
11 | 141aa45478673 |
12 | 59b560a06548 |
13 | 2555b3aa189b |
14 | 10d8db8d7b04 |
15 | 77a0c1a1615 |
hex | 3f07cefa388 |
4331423114120 has 32 divisors (see below), whose sum is σ = 9746022928320. Its totient is φ = 1732512192960.
The previous prime is 4331423114107. The next prime is 4331423114131. The reversal of 4331423114120 is 214113241334.
It is a super-2 number, since 2×43314231141202 (a number of 26 digits) contains 22 as substring.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 542342 + ... + 2992821.
It is an arithmetic number, because the mean of its divisors is an integer number (304563216510).
Almost surely, 24331423114120 is an apocalyptic number.
4331423114120 is a gapful number since it is divisible by the number (40) formed by its first and last digit.
It is an amenable number.
4331423114120 is an abundant number, since it is smaller than the sum of its proper divisors (5414599814200).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
4331423114120 is a wasteful number, since it uses less digits than its factorization.
4331423114120 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 3565805 (or 3565801 counting only the distinct ones).
The product of its (nonzero) digits is 6912, while the sum is 29.
Adding to 4331423114120 its reverse (214113241334), we get a palindrome (4545536355454).
The spelling of 4331423114120 in words is "four trillion, three hundred thirty-one billion, four hundred twenty-three million, one hundred fourteen thousand, one hundred twenty".
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