Base | Representation |
---|---|
bin | 1011001000001000100111… |
… | …1101101101011001101111 |
3 | 1121022121002002020200102021 |
4 | 2302002021331231121233 |
5 | 3100422003131424401 |
6 | 42004221451211011 |
7 | 2401622244012463 |
oct | 262021175553157 |
9 | 47277062220367 |
10 | 12234381514351 |
11 | 3997632462677 |
12 | 1457131746a67 |
13 | 6a990aa63a99 |
14 | 30420bdb04a3 |
15 | 16339e9a52a1 |
hex | b2089f6d66f |
12234381514351 has 2 divisors, whose sum is σ = 12234381514352. Its totient is φ = 12234381514350.
The previous prime is 12234381514283. The next prime is 12234381514361. The reversal of 12234381514351 is 15341518343221.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 12234381514351 - 213 = 12234381506159 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 12234381514298 and 12234381514307.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12234381514361) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6117190757175 + 6117190757176.
It is an arithmetic number, because the mean of its divisors is an integer number (6117190757176).
Almost surely, 212234381514351 is an apocalyptic number.
12234381514351 is a deficient number, since it is larger than the sum of its proper divisors (1).
12234381514351 is an equidigital number, since it uses as much as digits as its factorization.
12234381514351 is an evil number, because the sum of its binary digits is even.
The product of its digits is 345600, while the sum is 43.
Adding to 12234381514351 its reverse (15341518343221), we get a palindrome (27575899857572).
The spelling of 12234381514351 in words is "twelve trillion, two hundred thirty-four billion, three hundred eighty-one million, five hundred fourteen thousand, three hundred fifty-one".
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