Base | Representation |
---|---|
bin | 10100100011001001001… |
… | …001100101001111010111 |
3 | 11222222221100110002011111 |
4 | 110203021021211033113 |
5 | 141114012221132320 |
6 | 3000415410115451 |
7 | 204010514666251 |
oct | 24431111451727 |
9 | 4888840402144 |
10 | 1412124005335 |
11 | 4a4973206a0a |
12 | 1a981957bb87 |
13 | a3216668724 |
14 | 4c4c08a6cd1 |
15 | 26aec727a5a |
hex | 148c92653d7 |
1412124005335 has 8 divisors (see below), whose sum is σ = 1695183247296. Its totient is φ = 1129276243680.
The previous prime is 1412124005329. The next prime is 1412124005381. The reversal of 1412124005335 is 5335004212141.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 1412124005335 - 217 = 1412123874263 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 1412124005297 and 1412124005306.
It is a congruent number.
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, 52855384 + ... + 52882093.
It is an arithmetic number, because the mean of its divisors is an integer number (211897905912).
Almost surely, 21412124005335 is an apocalyptic number.
1412124005335 is a deficient number, since it is larger than the sum of its proper divisors (283059241961).
1412124005335 is a wasteful number, since it uses less digits than its factorization.
1412124005335 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 105740153.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 1412124005335 its reverse (5335004212141), we get a palindrome (6747128217476).
The spelling of 1412124005335 in words is "one trillion, four hundred twelve billion, one hundred twenty-four million, five thousand, three hundred thirty-five".
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