Base | Representation |
---|---|
bin | 11001110010010110100010… |
… | …101000101110000100101111 |
3 | 112212120000111021122212020021 |
4 | 121302112202220232010233 |
5 | 104331112411022044401 |
6 | 1041112241135503011 |
7 | 32613461505420412 |
oct | 3162264250560457 |
9 | 485500437585207 |
10 | 113411340034351 |
11 | 3315552a904987 |
12 | 10877a64387a67 |
13 | 4b38849941198 |
14 | 20011d1101b79 |
15 | d1a1511aa2a1 |
hex | 6725a2a2e12f |
113411340034351 has 2 divisors, whose sum is σ = 113411340034352. Its totient is φ = 113411340034350.
The previous prime is 113411340034229. The next prime is 113411340034367. The reversal of 113411340034351 is 153430043114311.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 113411340034351 - 223 = 113411331645743 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113411340034651) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56705670017175 + 56705670017176.
It is an arithmetic number, because the mean of its divisors is an integer number (56705670017176).
Almost surely, 2113411340034351 is an apocalyptic number.
113411340034351 is a deficient number, since it is larger than the sum of its proper divisors (1).
113411340034351 is an equidigital number, since it uses as much as digits as its factorization.
113411340034351 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 25920, while the sum is 34.
Adding to 113411340034351 its reverse (153430043114311), we get a palindrome (266841383148662).
The spelling of 113411340034351 in words is "one hundred thirteen trillion, four hundred eleven billion, three hundred forty million, thirty-four thousand, three hundred fifty-one".
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