Base | Representation |
---|---|
bin | 10000100101101111010… |
… | …101111000100111000101 |
3 | 11000222121221202000122212 |
4 | 100211233111320213011 |
5 | 122134242242410232 |
6 | 2231420300233205 |
7 | 145236054004544 |
oct | 20455725704705 |
9 | 4028557660585 |
10 | 1140034341317 |
11 | 3aa538735193 |
12 | 164b4323b805 |
13 | 83673cc2831 |
14 | 3d26c45415b |
15 | 1e9c54bceb2 |
hex | 1096f5789c5 |
1140034341317 has 2 divisors, whose sum is σ = 1140034341318. Its totient is φ = 1140034341316.
The previous prime is 1140034341233. The next prime is 1140034341359. The reversal of 1140034341317 is 7131434300411.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 866595289921 + 273439051396 = 930911^2 + 522914^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1140034341317 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1140034341367) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 570017170658 + 570017170659.
It is an arithmetic number, because the mean of its divisors is an integer number (570017170659).
Almost surely, 21140034341317 is an apocalyptic number.
It is an amenable number.
1140034341317 is a deficient number, since it is larger than the sum of its proper divisors (1).
1140034341317 is an equidigital number, since it uses as much as digits as its factorization.
1140034341317 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 12096, while the sum is 32.
Adding to 1140034341317 its reverse (7131434300411), we get a palindrome (8271468641728).
The spelling of 1140034341317 in words is "one trillion, one hundred forty billion, thirty-four million, three hundred forty-one thousand, three hundred seventeen".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.064 sec. • engine limits •