Base | Representation |
---|---|
bin | 1011111011010101111010… |
… | …0010010101010001110111 |
3 | 1201102200211122200211010211 |
4 | 2332311132202111101313 |
5 | 3204330210012304142 |
6 | 43520313310130251 |
7 | 2522315102041261 |
oct | 276653642252167 |
9 | 51380748624124 |
10 | 13114121213047 |
11 | 41a67385a41a8 |
12 | 15797307b0987 |
13 | 74186bc36b0a |
14 | 334a266b2b31 |
15 | 17b1dd499917 |
hex | bed5e895477 |
13114121213047 has 2 divisors, whose sum is σ = 13114121213048. Its totient is φ = 13114121213046.
The previous prime is 13114121212981. The next prime is 13114121213101. The reversal of 13114121213047 is 74031212141131.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13114121213047 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13114121913047) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6557060606523 + 6557060606524.
It is an arithmetic number, because the mean of its divisors is an integer number (6557060606524).
Almost surely, 213114121213047 is an apocalyptic number.
13114121213047 is a deficient number, since it is larger than the sum of its proper divisors (1).
13114121213047 is an equidigital number, since it uses as much as digits as its factorization.
13114121213047 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4032, while the sum is 31.
Adding to 13114121213047 its reverse (74031212141131), we get a palindrome (87145333354178).
The spelling of 13114121213047 in words is "thirteen trillion, one hundred fourteen billion, one hundred twenty-one million, two hundred thirteen thousand, forty-seven".
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