Base | Representation |
---|---|
bin | 10011000111110011101… |
… | …010111010111010101110 |
3 | 11122121210121210221101002 |
4 | 103013303222322322232 |
5 | 133012140100441032 |
6 | 2443400115120302 |
7 | 163636314253421 |
oct | 23076352727256 |
9 | 4577717727332 |
10 | 1314053140142 |
11 | 4673178163a6 |
12 | 1928098a3092 |
13 | 96bb7720a92 |
14 | 47859b359b8 |
15 | 242aca53862 |
hex | 131f3abaeae |
1314053140142 has 4 divisors (see below), whose sum is σ = 1971079710216. Its totient is φ = 657026570070.
The previous prime is 1314053140097. The next prime is 1314053140151. The reversal of 1314053140142 is 2410413504131.
It is a semiprime because it is the product of two primes.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 328513285034 + ... + 328513285037.
It is an arithmetic number, because the mean of its divisors is an integer number (492769927554).
Almost surely, 21314053140142 is an apocalyptic number.
1314053140142 is a deficient number, since it is larger than the sum of its proper divisors (657026570074).
1314053140142 is an equidigital number, since it uses as much as digits as its factorization.
1314053140142 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 657026570073.
The product of its (nonzero) digits is 5760, while the sum is 29.
Adding to 1314053140142 its reverse (2410413504131), we get a palindrome (3724466644273).
The spelling of 1314053140142 in words is "one trillion, three hundred fourteen billion, fifty-three million, one hundred forty thousand, one hundred forty-two".
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