Base | Representation |
---|---|
bin | 11101110101101001101010… |
… | …000111011101001100000111 |
3 | 122012122111010001020100212111 |
4 | 131311221222013131030013 |
5 | 114200033433020221242 |
6 | 1143034154323201451 |
7 | 36433032652425463 |
oct | 3565515207351407 |
9 | 565574101210774 |
10 | 131230211101447 |
11 | 388a5469002a57 |
12 | 12875364216287 |
13 | 582bc56ba8283 |
14 | 245980cba34a3 |
15 | 10288eb6ab817 |
hex | 775a6a1dd307 |
131230211101447 has 2 divisors, whose sum is σ = 131230211101448. Its totient is φ = 131230211101446.
The previous prime is 131230211101433. The next prime is 131230211101459. The reversal of 131230211101447 is 744101112032131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131230211101447 - 243 = 122434118079239 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131230211101417) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65615105550723 + 65615105550724.
It is an arithmetic number, because the mean of its divisors is an integer number (65615105550724).
Almost surely, 2131230211101447 is an apocalyptic number.
131230211101447 is a deficient number, since it is larger than the sum of its proper divisors (1).
131230211101447 is an equidigital number, since it uses as much as digits as its factorization.
131230211101447 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 131230211101447 its reverse (744101112032131), we get a palindrome (875331323133578).
The spelling of 131230211101447 in words is "one hundred thirty-one trillion, two hundred thirty billion, two hundred eleven million, one hundred one thousand, four hundred forty-seven".
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