Base | Representation |
---|---|
bin | 10011101010010100011… |
… | …011111001001101011011 |
3 | 11210011110011222002011121 |
4 | 103222110123321031123 |
5 | 134114033134234011 |
6 | 2512405201510111 |
7 | 166420524054202 |
oct | 23522433711533 |
9 | 4704404862147 |
10 | 1351110071131 |
11 | 4810034590a7 |
12 | 199a30042337 |
13 | 9a541b32b92 |
14 | 495734ba239 |
15 | 2522ae8ee71 |
hex | 13a946f935b |
1351110071131 has 2 divisors, whose sum is σ = 1351110071132. Its totient is φ = 1351110071130.
The previous prime is 1351110071117. The next prime is 1351110071159. The reversal of 1351110071131 is 1311700111531.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1351110071131 - 233 = 1342520136539 is a prime.
It is a super-2 number, since 2×13511100711312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1351110071096 and 1351110071105.
It is not a weakly prime, because it can be changed into another prime (1351110071431) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 675555035565 + 675555035566.
It is an arithmetic number, because the mean of its divisors is an integer number (675555035566).
Almost surely, 21351110071131 is an apocalyptic number.
1351110071131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1351110071131 is an equidigital number, since it uses as much as digits as its factorization.
1351110071131 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 315, while the sum is 25.
Adding to 1351110071131 its reverse (1311700111531), we get a palindrome (2662810182662).
The spelling of 1351110071131 in words is "one trillion, three hundred fifty-one billion, one hundred ten million, seventy-one thousand, one hundred thirty-one".
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