Base | Representation |
---|---|
bin | 11101110101101100100100… |
… | …110110101001001001011011 |
3 | 122012122210012101101210112012 |
4 | 131311230210312221021123 |
5 | 114200111342033031441 |
6 | 1143035433240540135 |
7 | 36433203414053006 |
oct | 3565544466511133 |
9 | 565583171353465 |
10 | 131233344033371 |
11 | 388a6826508098 |
12 | 12875a9948164b |
13 | 582c335c93948 |
14 | 2459a28cd053d |
15 | 1028a3175c9eb |
hex | 775b24da925b |
131233344033371 has 2 divisors, whose sum is σ = 131233344033372. Its totient is φ = 131233344033370.
The previous prime is 131233344033329. The next prime is 131233344033413. The reversal of 131233344033371 is 173330443332131.
It is a balanced prime because it is at equal distance from previous prime (131233344033329) and next prime (131233344033413).
It is a cyclic number.
It is not a de Polignac number, because 131233344033371 - 26 = 131233344033307 is a prime.
It is a super-2 number, since 2×1312333440333712 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131233344003371) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65616672016685 + 65616672016686.
It is an arithmetic number, because the mean of its divisors is an integer number (65616672016686).
Almost surely, 2131233344033371 is an apocalyptic number.
131233344033371 is a deficient number, since it is larger than the sum of its proper divisors (1).
131233344033371 is an equidigital number, since it uses as much as digits as its factorization.
131233344033371 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 489888, while the sum is 41.
The spelling of 131233344033371 in words is "one hundred thirty-one trillion, two hundred thirty-three billion, three hundred forty-four million, thirty-three thousand, three hundred seventy-one".
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