Base | Representation |
---|---|
bin | 100101001101011101000… |
… | …1001111010100001111011 |
3 | 200002220110210211102122002 |
4 | 1022122322021322201323 |
5 | 1132242221331242301 |
6 | 14513222325251215 |
7 | 1035325004552402 |
oct | 112327211724173 |
9 | 20086423742562 |
10 | 5114134243451 |
11 | 16a198a5009a2 |
12 | 6a71a1263b0b |
13 | 2b1350c5596c |
14 | 13974da66639 |
15 | 8d06c75a16b |
hex | 4a6ba27a87b |
5114134243451 has 2 divisors, whose sum is σ = 5114134243452. Its totient is φ = 5114134243450.
The previous prime is 5114134243423. The next prime is 5114134243453. The reversal of 5114134243451 is 1543424314115.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5114134243451 - 238 = 4839256336507 is a prime.
Together with 5114134243453, 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 (5114134243453) 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, 2557067121725 + 2557067121726.
It is an arithmetic number, because the mean of its divisors is an integer number (2557067121726).
Almost surely, 25114134243451 is an apocalyptic number.
5114134243451 is a deficient number, since it is larger than the sum of its proper divisors (1).
5114134243451 is an equidigital number, since it uses as much as digits as its factorization.
5114134243451 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 115200, while the sum is 38.
Adding to 5114134243451 its reverse (1543424314115), we get a palindrome (6657558557566).
The spelling of 5114134243451 in words is "five trillion, one hundred fourteen billion, one hundred thirty-four million, two hundred forty-three thousand, four hundred fifty-one".
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