Base | Representation |
---|---|
bin | 11001100001000001001001… |
… | …010001110011010101000011 |
3 | 112201100010210011001220010221 |
4 | 121201001021101303111003 |
5 | 104202103314014211342 |
6 | 1034401115052003511 |
7 | 32431431362102116 |
oct | 3141011121632503 |
9 | 481303704056127 |
10 | 112220134913347 |
11 | 3283632364962a |
12 | 10705020160597 |
13 | 4a8040c4785b6 |
14 | 1d9d6aaa7657d |
15 | ce918367ea67 |
hex | 661049473543 |
112220134913347 has 2 divisors, whose sum is σ = 112220134913348. Its totient is φ = 112220134913346.
The previous prime is 112220134913329. The next prime is 112220134913351. The reversal of 112220134913347 is 743319431022211.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 112220134913347 - 223 = 112220126524739 is a prime.
It is not a weakly prime, because it can be changed into another prime (112220134913647) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56110067456673 + 56110067456674.
It is an arithmetic number, because the mean of its divisors is an integer number (56110067456674).
Almost surely, 2112220134913347 is an apocalyptic number.
112220134913347 is a deficient number, since it is larger than the sum of its proper divisors (1).
112220134913347 is an equidigital number, since it uses as much as digits as its factorization.
112220134913347 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 217728, while the sum is 43.
Adding to 112220134913347 its reverse (743319431022211), we get a palindrome (855539565935558).
The spelling of 112220134913347 in words is "one hundred twelve trillion, two hundred twenty billion, one hundred thirty-four million, nine hundred thirteen thousand, three hundred forty-seven".
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