Base | Representation |
---|---|
bin | 11110000000110101110… |
… | …01100010000000010111 |
3 | 10122120211010110122002002 |
4 | 33000122321202000113 |
5 | 113343441310041101 |
6 | 2105425141505515 |
7 | 134335113416561 |
oct | 17003271420027 |
9 | 3576733418062 |
10 | 1031243440151 |
11 | 368391045a11 |
12 | 147a4141229b |
13 | 76327154732 |
14 | 37caba8c331 |
15 | 1bc5960bd6b |
hex | f01ae62017 |
1031243440151 has 2 divisors, whose sum is σ = 1031243440152. Its totient is φ = 1031243440150.
The previous prime is 1031243440121. The next prime is 1031243440169. The reversal of 1031243440151 is 1510443421301.
It is a strong prime.
It is an emirp because it is prime and its reverse (1510443421301) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1031243440151 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1031243440111) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 515621720075 + 515621720076.
It is an arithmetic number, because the mean of its divisors is an integer number (515621720076).
Almost surely, 21031243440151 is an apocalyptic number.
1031243440151 is a deficient number, since it is larger than the sum of its proper divisors (1).
1031243440151 is an equidigital number, since it uses as much as digits as its factorization.
1031243440151 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5760, while the sum is 29.
Adding to 1031243440151 its reverse (1510443421301), we get a palindrome (2541686861452).
The spelling of 1031243440151 in words is "one trillion, thirty-one billion, two hundred forty-three million, four hundred forty thousand, one hundred fifty-one".
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