Base | Representation |
---|---|
bin | 10010111100101011110… |
… | …111101110101010010001 |
3 | 11121110222100210110110102 |
4 | 102330223313232222101 |
5 | 132313210421320032 |
6 | 2434103120142145 |
7 | 163034345621261 |
oct | 22745367565221 |
9 | 4543870713412 |
10 | 1302111120017 |
11 | 462249865984 |
12 | 190436467955 |
13 | 95a335a9896 |
14 | 47045acb7a1 |
15 | 23d0e437362 |
hex | 12f2bdeea91 |
1302111120017 has 2 divisors, whose sum is σ = 1302111120018. Its totient is φ = 1302111120016.
The previous prime is 1302111119993. The next prime is 1302111120019. The reversal of 1302111120017 is 7100211112031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1110924432016 + 191186688001 = 1054004^2 + 437249^2 .
It is a cyclic number.
It is not a de Polignac number, because 1302111120017 - 234 = 1284931250833 is a prime.
Together with 1302111120019, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1302111120019) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 651055560008 + 651055560009.
It is an arithmetic number, because the mean of its divisors is an integer number (651055560009).
Almost surely, 21302111120017 is an apocalyptic number.
It is an amenable number.
1302111120017 is a deficient number, since it is larger than the sum of its proper divisors (1).
1302111120017 is an equidigital number, since it uses as much as digits as its factorization.
1302111120017 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 84, while the sum is 20.
Adding to 1302111120017 its reverse (7100211112031), we get a palindrome (8402322232048).
The spelling of 1302111120017 in words is "one trillion, three hundred two billion, one hundred eleven million, one hundred twenty thousand, seventeen".
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