Base | Representation |
---|---|
bin | 11101110100011000100011… |
… | …111011111010001100001111 |
3 | 122012100011100111220201111112 |
4 | 131310120203323322030033 |
5 | 114142122114341322341 |
6 | 1142530154005442235 |
7 | 36423531060141224 |
oct | 3564304373721417 |
9 | 565304314821445 |
10 | 131143134323471 |
11 | 38871544481936 |
12 | 1286050241037b |
13 | 582398a89788a |
14 | 245550c20394b |
15 | 1026501ca9aeb |
hex | 774623efa30f |
131143134323471 has 2 divisors, whose sum is σ = 131143134323472. Its totient is φ = 131143134323470.
The previous prime is 131143134323449. The next prime is 131143134323489. The reversal of 131143134323471 is 174323431341131.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131143134323471 is a prime.
It is a super-3 number, since 3×1311431343234713 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131143134323371) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65571567161735 + 65571567161736.
It is an arithmetic number, because the mean of its divisors is an integer number (65571567161736).
Almost surely, 2131143134323471 is an apocalyptic number.
131143134323471 is a deficient number, since it is larger than the sum of its proper divisors (1).
131143134323471 is an equidigital number, since it uses as much as digits as its factorization.
131143134323471 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 217728, while the sum is 41.
The spelling of 131143134323471 in words is "one hundred thirty-one trillion, one hundred forty-three billion, one hundred thirty-four million, three hundred twenty-three thousand, four hundred seventy-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.074 sec. • engine limits •