Base | Representation |
---|---|
bin | 101111101000110010… |
… | …001001101001111001 |
3 | 11220000201221100202101 |
4 | 233220302021221321 |
5 | 1314223411213423 |
6 | 35255324103401 |
7 | 3460355454136 |
oct | 575062115171 |
9 | 156021840671 |
10 | 51150101113 |
11 | 1a768998041 |
12 | 9ab6094b61 |
13 | 4a92112b27 |
14 | 269339158d |
15 | 14e58257ad |
hex | be8c89a79 |
51150101113 has 2 divisors, whose sum is σ = 51150101114. Its totient is φ = 51150101112.
The previous prime is 51150101059. The next prime is 51150101161. The reversal of 51150101113 is 31110105115.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 50098525929 + 1051575184 = 223827^2 + 32428^2 .
It is a cyclic number.
It is not a de Polignac number, because 51150101113 - 221 = 51148003961 is a prime.
It is not a weakly prime, because it can be changed into another prime (51150101213) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25575050556 + 25575050557.
It is an arithmetic number, because the mean of its divisors is an integer number (25575050557).
Almost surely, 251150101113 is an apocalyptic number.
It is an amenable number.
51150101113 is a deficient number, since it is larger than the sum of its proper divisors (1).
51150101113 is an equidigital number, since it uses as much as digits as its factorization.
51150101113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 75, while the sum is 19.
Adding to 51150101113 its reverse (31110105115), we get a palindrome (82260206228).
The spelling of 51150101113 in words is "fifty-one billion, one hundred fifty million, one hundred one thousand, one hundred thirteen".
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