Base | Representation |
---|---|
bin | 10100100011011001101… |
… | …010101010110010111111 |
3 | 12000000122122212200012121 |
4 | 110203121222222302333 |
5 | 141120044141124203 |
6 | 3000503105343411 |
7 | 204020421246154 |
oct | 24433152526277 |
9 | 5000578780177 |
10 | 1412401114303 |
11 | 4a4aa5675a93 |
12 | 1a9896337b67 |
13 | a325bbb022a |
14 | 4c5095dc22b |
15 | 26b16c140bd |
hex | 148d9aaacbf |
1412401114303 has 4 divisors (see below), whose sum is σ = 1412436479328. Its totient is φ = 1412365749280.
The previous prime is 1412401114259. The next prime is 1412401114321. The reversal of 1412401114303 is 3034111042141.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1412401114303 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1412401114903) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 17622538 + ... + 17702503.
It is an arithmetic number, because the mean of its divisors is an integer number (353109119832).
Almost surely, 21412401114303 is an apocalyptic number.
1412401114303 is a deficient number, since it is larger than the sum of its proper divisors (35365025).
1412401114303 is an equidigital number, since it uses as much as digits as its factorization.
1412401114303 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 35365024.
The product of its (nonzero) digits is 1152, while the sum is 25.
Adding to 1412401114303 its reverse (3034111042141), we get a palindrome (4446512156444).
The spelling of 1412401114303 in words is "one trillion, four hundred twelve billion, four hundred one million, one hundred fourteen thousand, three hundred three".
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