Base | Representation |
---|---|
bin | 11001110100001110011001… |
… | …111001110100010100111011 |
3 | 112220000022200210010120200121 |
4 | 121310032121321310110323 |
5 | 104340220001342242342 |
6 | 1041251324145114111 |
7 | 32625666045602413 |
oct | 3164163171642473 |
9 | 486008623116617 |
10 | 113540042540347 |
11 | 331a5075038615 |
12 | 108989a2628937 |
13 | 4b47a1aa77587 |
14 | 200752013d643 |
15 | d1d685176067 |
hex | 674399e7453b |
113540042540347 has 2 divisors, whose sum is σ = 113540042540348. Its totient is φ = 113540042540346.
The previous prime is 113540042540279. The next prime is 113540042540377. The reversal of 113540042540347 is 743045240045311.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-113540042540347 is a prime.
It is a super-2 number, since 2×1135400425403472 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (113540042540377) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56770021270173 + 56770021270174.
It is an arithmetic number, because the mean of its divisors is an integer number (56770021270174).
Almost surely, 2113540042540347 is an apocalyptic number.
113540042540347 is a deficient number, since it is larger than the sum of its proper divisors (1).
113540042540347 is an equidigital number, since it uses as much as digits as its factorization.
113540042540347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 806400, while the sum is 43.
Adding to 113540042540347 its reverse (743045240045311), we get a palindrome (856585282585658).
The spelling of 113540042540347 in words is "one hundred thirteen trillion, five hundred forty billion, forty-two million, five hundred forty thousand, three hundred forty-seven".
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