Base | Representation |
---|---|
bin | 110101110010001100… |
… | …0000000111010000101 |
3 | 102001010102211221210012 |
4 | 1223210120000322011 |
5 | 3343021131300432 |
6 | 125021003440005 |
7 | 11226133220156 |
oct | 1534430007205 |
9 | 361112757705 |
10 | 115500650117 |
11 | 44a90118a46 |
12 | 1a474b26005 |
13 | ab78caa768 |
14 | 583996992d |
15 | 300ee94cb2 |
hex | 1ae4600e85 |
115500650117 has 2 divisors, whose sum is σ = 115500650118. Its totient is φ = 115500650116.
The previous prime is 115500650099. The next prime is 115500650137. The reversal of 115500650117 is 711056005511.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 68435083201 + 47065566916 = 261601^2 + 216946^2 .
It is a cyclic number.
It is not a de Polignac number, because 115500650117 - 224 = 115483872901 is a prime.
It is a super-2 number, since 2×1155006501172 (a number of 23 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (115500650137) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57750325058 + 57750325059.
It is an arithmetic number, because the mean of its divisors is an integer number (57750325059).
Almost surely, 2115500650117 is an apocalyptic number.
It is an amenable number.
115500650117 is a deficient number, since it is larger than the sum of its proper divisors (1).
115500650117 is an equidigital number, since it uses as much as digits as its factorization.
115500650117 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5250, while the sum is 32.
Adding to 115500650117 its reverse (711056005511), we get a palindrome (826556655628).
The spelling of 115500650117 in words is "one hundred fifteen billion, five hundred million, six hundred fifty thousand, one hundred seventeen".
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