Base | Representation |
---|---|
bin | 11101110111010011000111… |
… | …010100110001110100110111 |
3 | 122020001022101102221212010021 |
4 | 131313103013110301310313 |
5 | 114203412333132242011 |
6 | 1143202202321510011 |
7 | 36444151661411116 |
oct | 3567230724616467 |
9 | 566038342855107 |
10 | 131343444024631 |
11 | 38939495043103 |
12 | 128932a57a2307 |
13 | 5839831100377 |
14 | 24610b14a1c7d |
15 | 102b827507471 |
hex | 7774c7531d37 |
131343444024631 has 2 divisors, whose sum is σ = 131343444024632. Its totient is φ = 131343444024630.
The previous prime is 131343444024601. The next prime is 131343444024647. The reversal of 131343444024631 is 136420444343131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131343444024631 - 25 = 131343444024599 is a prime.
It is a super-2 number, since 2×1313434440246312 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131343444024601) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65671722012315 + 65671722012316.
It is an arithmetic number, because the mean of its divisors is an integer number (65671722012316).
Almost surely, 2131343444024631 is an apocalyptic number.
131343444024631 is a deficient number, since it is larger than the sum of its proper divisors (1).
131343444024631 is an equidigital number, since it uses as much as digits as its factorization.
131343444024631 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 995328, while the sum is 43.
Adding to 131343444024631 its reverse (136420444343131), we get a palindrome (267763888367762).
The spelling of 131343444024631 in words is "one hundred thirty-one trillion, three hundred forty-three billion, four hundred forty-four million, twenty-four thousand, six hundred thirty-one".
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