Base | Representation |
---|---|
bin | 11001101100011001100000… |
… | …101110111101100110100111 |
3 | 112211002220110110222202112111 |
4 | 121230121200232331212213 |
5 | 104302412012343401242 |
6 | 1040200255540301451 |
7 | 32542066111536112 |
oct | 3154314056754647 |
9 | 484086413882474 |
10 | 113002212481447 |
11 | 33007a72a90a33 |
12 | 108107043a1887 |
13 | 4b090a82b2b83 |
14 | 1dc949ca03579 |
15 | d0e6a82db817 |
hex | 66c660bbd9a7 |
113002212481447 has 2 divisors, whose sum is σ = 113002212481448. Its totient is φ = 113002212481446.
The previous prime is 113002212481351. The next prime is 113002212481451. The reversal of 113002212481447 is 744184212200311.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-113002212481447 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113002212481247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56501106240723 + 56501106240724.
It is an arithmetic number, because the mean of its divisors is an integer number (56501106240724).
Almost surely, 2113002212481447 is an apocalyptic number.
113002212481447 is a deficient number, since it is larger than the sum of its proper divisors (1).
113002212481447 is an equidigital number, since it uses as much as digits as its factorization.
113002212481447 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 86016, while the sum is 40.
Adding to 113002212481447 its reverse (744184212200311), we get a palindrome (857186424681758).
The spelling of 113002212481447 in words is "one hundred thirteen trillion, two billion, two hundred twelve million, four hundred eighty-one thousand, four hundred forty-seven".
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