Base | Representation |
---|---|
bin | 10100100011000111111… |
… | …111011010001011000111 |
3 | 11222222212222212020000221 |
4 | 110203013333122023013 |
5 | 141113442231430401 |
6 | 3000413433245211 |
7 | 204010160504251 |
oct | 24430777321307 |
9 | 4888788766027 |
10 | 1412104561351 |
11 | 4a4963236400 |
12 | 1a9812b63807 |
13 | a321261b403 |
14 | 4c4bc084cd1 |
15 | 26aeab867a1 |
hex | 148c7fda2c7 |
1412104561351 has 24 divisors (see below), whose sum is σ = 1568919889152. Its totient is φ = 1269872947200.
The previous prime is 1412104561297. The next prime is 1412104561363. The reversal of 1412104561351 is 1531654012141.
It is not a de Polignac number, because 1412104561351 - 27 = 1412104561223 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1412104561951) 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 23 ways as a sum of consecutive naturals, for example, 24748311 + ... + 24805303.
It is an arithmetic number, because the mean of its divisors is an integer number (65371662048).
Almost surely, 21412104561351 is an apocalyptic number.
1412104561351 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
1412104561351 is a deficient number, since it is larger than the sum of its proper divisors (156815327801).
1412104561351 is a wasteful number, since it uses less digits than its factorization.
1412104561351 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 59223 (or 59212 counting only the distinct ones).
The product of its (nonzero) digits is 14400, while the sum is 34.
Adding to 1412104561351 its reverse (1531654012141), we get a palindrome (2943758573492).
The spelling of 1412104561351 in words is "one trillion, four hundred twelve billion, one hundred four million, five hundred sixty-one thousand, three hundred fifty-one".
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