Base | Representation |
---|---|
bin | 10011100001000000100… |
… | …001000011110000001111 |
3 | 11202012122102012211012121 |
4 | 103201000201003300033 |
5 | 133433044211014203 |
6 | 2504033132513411 |
7 | 165615003435334 |
oct | 23410041036017 |
9 | 4665572184177 |
10 | 1341112204303 |
11 | 477842970843 |
12 | 197abb91b867 |
13 | 9960a711651 |
14 | 48ca575098b |
15 | 24d433a40bd |
hex | 13840843c0f |
1341112204303 has 2 divisors, whose sum is σ = 1341112204304. Its totient is φ = 1341112204302.
The previous prime is 1341112204241. The next prime is 1341112204363. The reversal of 1341112204303 is 3034022111431.
It is a strong prime.
It is an emirp because it is prime and its reverse (3034022111431) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1341112204303 is a prime.
It is a super-2 number, since 2×13411122043032 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1341112204363) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670556102151 + 670556102152.
It is an arithmetic number, because the mean of its divisors is an integer number (670556102152).
Almost surely, 21341112204303 is an apocalyptic number.
1341112204303 is a deficient number, since it is larger than the sum of its proper divisors (1).
1341112204303 is an equidigital number, since it uses as much as digits as its factorization.
1341112204303 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1728, while the sum is 25.
Adding to 1341112204303 its reverse (3034022111431), we get a palindrome (4375134315734).
The spelling of 1341112204303 in words is "one trillion, three hundred forty-one billion, one hundred twelve million, two hundred four thousand, three hundred three".
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