Base | Representation |
---|---|
bin | 11001110001001011000100… |
… | …010001100011111010010011 |
3 | 112212021021022110120002012202 |
4 | 121301023010101203322103 |
5 | 104323300423300311011 |
6 | 1041011115430513415 |
7 | 32604562324526261 |
oct | 3161130421437223 |
9 | 485237273502182 |
10 | 113330300010131 |
11 | 33124123928256 |
12 | 1086420820586b |
13 | 4b310031c7629 |
14 | 1ddb30419b431 |
15 | d17eab75123b |
hex | 6712c4463e93 |
113330300010131 has 2 divisors, whose sum is σ = 113330300010132. Its totient is φ = 113330300010130.
The previous prime is 113330300010101. The next prime is 113330300010163. The reversal of 113330300010131 is 131010003033311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 113330300010131 - 214 = 113330299993747 is a prime.
It is a super-3 number, since 3×1133303000101313 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (113330300010101) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56665150005065 + 56665150005066.
It is an arithmetic number, because the mean of its divisors is an integer number (56665150005066).
Almost surely, 2113330300010131 is an apocalyptic number.
113330300010131 is a deficient number, since it is larger than the sum of its proper divisors (1).
113330300010131 is an equidigital number, since it uses as much as digits as its factorization.
113330300010131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 243, while the sum is 20.
Adding to 113330300010131 its reverse (131010003033311), we get a palindrome (244340303043442).
The spelling of 113330300010131 in words is "one hundred thirteen trillion, three hundred thirty billion, three hundred million, ten thousand, one hundred thirty-one".
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