Base | Representation |
---|---|
bin | 1001010100000100… |
… | …0111111111010011 |
3 | 20110020101020112012 |
4 | 2111001013333103 |
5 | 20110011200201 |
6 | 1052025453135 |
7 | 115645333301 |
oct | 22501077723 |
9 | 6406336465 |
10 | 2500100051 |
11 | 1073272199 |
12 | 5993411ab |
13 | 30ac65b57 |
14 | 19a079271 |
15 | e974a5bb |
hex | 95047fd3 |
2500100051 has 2 divisors, whose sum is σ = 2500100052. Its totient is φ = 2500100050.
The previous prime is 2500100017. The next prime is 2500100111. The reversal of 2500100051 is 1500010052.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2500100051 - 214 = 2500083667 is a prime.
It is a Sophie Germain prime.
It is a junction number, because it is equal to n+sod(n) for n = 2500099999 and 2500100035.
It is not a weakly prime, because it can be changed into another prime (2500180051) 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, 1250050025 + 1250050026.
It is an arithmetic number, because the mean of its divisors is an integer number (1250050026).
Almost surely, 22500100051 is an apocalyptic number.
2500100051 is a deficient number, since it is larger than the sum of its proper divisors (1).
2500100051 is an equidigital number, since it uses as much as digits as its factorization.
2500100051 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 50, while the sum is 14.
The square root of 2500100051 is about 50001.0004999900. Note that the first 3 decimals coincide. The cubic root of 2500100051 is about 1357.2269134024.
Subtracting from 2500100051 its product of nonzero digits (50), we obtain a square (2500100001 = 500012).
The spelling of 2500100051 in words is "two billion, five hundred million, one hundred thousand, fifty-one".
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