Base | Representation |
---|---|
bin | 10111011110110000010010… |
… | …11100010010100101010011 |
3 | 20202211012001000201102100011 |
4 | 23233120021130102211103 |
5 | 23231433423320243403 |
6 | 301452241022535351 |
7 | 13606310656662064 |
oct | 1357301134224523 |
9 | 222735030642304 |
10 | 51634255243603 |
11 | 154a7a67a8a05a |
12 | 595b093737557 |
13 | 22a712148cb16 |
14 | ca717b53656b |
15 | 5e81d3710a6d |
hex | 2ef609712953 |
51634255243603 has 2 divisors, whose sum is σ = 51634255243604. Its totient is φ = 51634255243602.
The previous prime is 51634255243561. The next prime is 51634255243619. The reversal of 51634255243603 is 30634255243615.
51634255243603 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-51634255243603 is a prime.
It is not a weakly prime, because it can be changed into another prime (51634255243663) 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, 25817127621801 + 25817127621802.
It is an arithmetic number, because the mean of its divisors is an integer number (25817127621802).
Almost surely, 251634255243603 is an apocalyptic number.
51634255243603 is a deficient number, since it is larger than the sum of its proper divisors (1).
51634255243603 is an equidigital number, since it uses as much as digits as its factorization.
51634255243603 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 7776000, while the sum is 49.
The spelling of 51634255243603 in words is "fifty-one trillion, six hundred thirty-four billion, two hundred fifty-five million, two hundred forty-three thousand, six hundred three".
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