Base | Representation |
---|---|
bin | 11101101000101100101000… |
… | …000010101111101110010101 |
3 | 122002111101102210102120012222 |
4 | 131220230220002233232111 |
5 | 114040442402322340432 |
6 | 1141113220104152125 |
7 | 36311520543000323 |
oct | 3550545002575625 |
9 | 562441383376188 |
10 | 130340044340117 |
11 | 385919a2087997 |
12 | 12750935168645 |
13 | 57960236bc4b1 |
14 | 24286c5730d13 |
15 | 101069c59c112 |
hex | 768b280afb95 |
130340044340117 has 2 divisors, whose sum is σ = 130340044340118. Its totient is φ = 130340044340116.
The previous prime is 130340044340063. The next prime is 130340044340221. The reversal of 130340044340117 is 711043440043031.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 67290505080241 + 63049539259876 = 8203079^2 + 7940374^2 .
It is a cyclic number.
It is not a de Polignac number, because 130340044340117 - 214 = 130340044323733 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130340044340017) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65170022170058 + 65170022170059.
It is an arithmetic number, because the mean of its divisors is an integer number (65170022170059).
Almost surely, 2130340044340117 is an apocalyptic number.
It is an amenable number.
130340044340117 is a deficient number, since it is larger than the sum of its proper divisors (1).
130340044340117 is an equidigital number, since it uses as much as digits as its factorization.
130340044340117 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 48384, while the sum is 35.
Adding to 130340044340117 its reverse (711043440043031), we get a palindrome (841383484383148).
The spelling of 130340044340117 in words is "one hundred thirty trillion, three hundred forty billion, forty-four million, three hundred forty thousand, one hundred seventeen".
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