Base | Representation |
---|---|
bin | 10100101011100001011… |
… | …000101011101000011011 |
3 | 12000212011100000222100012 |
4 | 110223201120223220123 |
5 | 141240423330121301 |
6 | 3004504225232135 |
7 | 204446456243546 |
oct | 24534130535033 |
9 | 5025140028305 |
10 | 1421120551451 |
11 | 4a876a559197 |
12 | 1ab50a49264b |
13 | a401b4a98b3 |
14 | 4cad565d85d |
15 | 26e77470ebb |
hex | 14ae162ba1b |
1421120551451 has 2 divisors, whose sum is σ = 1421120551452. Its totient is φ = 1421120551450.
The previous prime is 1421120551439. The next prime is 1421120551453. The reversal of 1421120551451 is 1541550211241.
It is a strong prime.
It is an emirp because it is prime and its reverse (1541550211241) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1421120551451 - 210 = 1421120550427 is a prime.
Together with 1421120551453, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1421120551453) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 710560275725 + 710560275726.
It is an arithmetic number, because the mean of its divisors is an integer number (710560275726).
Almost surely, 21421120551451 is an apocalyptic number.
1421120551451 is a deficient number, since it is larger than the sum of its proper divisors (1).
1421120551451 is an equidigital number, since it uses as much as digits as its factorization.
1421120551451 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8000, while the sum is 32.
Adding to 1421120551451 its reverse (1541550211241), we get a palindrome (2962670762692).
The spelling of 1421120551451 in words is "one trillion, four hundred twenty-one billion, one hundred twenty million, five hundred fifty-one thousand, four hundred fifty-one".
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