Base | Representation |
---|---|
bin | 10111011100011111100011… |
… | …111000100001101101100011 |
3 | 111112002111221110001200012121 |
4 | 113130133203320201231203 |
5 | 102003401113344340001 |
6 | 1003145301114405111 |
7 | 30501450551104456 |
oct | 2734374370415543 |
9 | 445074843050177 |
10 | 103113103121251 |
11 | 2a945026616983 |
12 | b694015730797 |
13 | 456c6a7021662 |
14 | 1b669b509949d |
15 | bdc31d76e6a1 |
hex | 5dc7e3e21b63 |
103113103121251 has 2 divisors, whose sum is σ = 103113103121252. Its totient is φ = 103113103121250.
The previous prime is 103113103121209. The next prime is 103113103121299. The reversal of 103113103121251 is 152121301311301.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 103113103121251 - 215 = 103113103088483 is a prime.
It is not a weakly prime, because it can be changed into another prime (103113103121551) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 51556551560625 + 51556551560626.
It is an arithmetic number, because the mean of its divisors is an integer number (51556551560626).
Almost surely, 2103113103121251 is an apocalyptic number.
103113103121251 is a deficient number, since it is larger than the sum of its proper divisors (1).
103113103121251 is an equidigital number, since it uses as much as digits as its factorization.
103113103121251 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 540, while the sum is 25.
Adding to 103113103121251 its reverse (152121301311301), we get a palindrome (255234404432552).
The spelling of 103113103121251 in words is "one hundred three trillion, one hundred thirteen billion, one hundred three million, one hundred twenty-one thousand, two hundred fifty-one".
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