Base | Representation |
---|---|
bin | 11101110101101001101010… |
… | …000111011101001011111001 |
3 | 122012122111010001020100211222 |
4 | 131311221222013131023321 |
5 | 114200033433020221213 |
6 | 1143034154323201425 |
7 | 36433032652425443 |
oct | 3565515207351371 |
9 | 565574101210758 |
10 | 131230211101433 |
11 | 388a5469002a44 |
12 | 12875364216275 |
13 | 582bc56ba8272 |
14 | 245980cba3493 |
15 | 10288eb6ab808 |
hex | 775a6a1dd2f9 |
131230211101433 has 2 divisors, whose sum is σ = 131230211101434. Its totient is φ = 131230211101432.
The previous prime is 131230211101417. The next prime is 131230211101447. The reversal of 131230211101433 is 334101112032131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 117050068585024 + 14180142516409 = 10818968^2 + 3765653^2 .
It is a cyclic number.
It is not a de Polignac number, because 131230211101433 - 24 = 131230211101417 is a prime.
It is not a weakly prime, because it can be changed into another prime (131230211301433) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65615105550716 + 65615105550717.
It is an arithmetic number, because the mean of its divisors is an integer number (65615105550717).
Almost surely, 2131230211101433 is an apocalyptic number.
It is an amenable number.
131230211101433 is a deficient number, since it is larger than the sum of its proper divisors (1).
131230211101433 is an equidigital number, since it uses as much as digits as its factorization.
131230211101433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1296, while the sum is 26.
Adding to 131230211101433 its reverse (334101112032131), we get a palindrome (465331323133564).
The spelling of 131230211101433 in words is "one hundred thirty-one trillion, two hundred thirty billion, two hundred eleven million, one hundred one thousand, four hundred thirty-three".
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