Base | Representation |
---|---|
bin | 10010111010111011001… |
… | …010001011101110000011 |
3 | 11121022002120120022212121 |
4 | 102322323022023232003 |
5 | 132300322323212402 |
6 | 2433151325505111 |
7 | 162636444525322 |
oct | 22727312135603 |
9 | 4538076508777 |
10 | 1300220132227 |
11 | 461469405855 |
12 | 18bba9112197 |
13 | 957c18b00b3 |
14 | 46d068cc0b9 |
15 | 23c4d409837 |
hex | 12ebb28bb83 |
1300220132227 has 2 divisors, whose sum is σ = 1300220132228. Its totient is φ = 1300220132226.
The previous prime is 1300220132149. The next prime is 1300220132327. The reversal of 1300220132227 is 7222310220031.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1300220132227 - 239 = 750464318339 is a prime.
It is a super-4 number, since 4×13002201322274 (a number of 50 digits) contains 4444 as substring.
It is not a weakly prime, because it can be changed into another prime (1300220132327) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650110066113 + 650110066114.
It is an arithmetic number, because the mean of its divisors is an integer number (650110066114).
Almost surely, 21300220132227 is an apocalyptic number.
1300220132227 is a deficient number, since it is larger than the sum of its proper divisors (1).
1300220132227 is an equidigital number, since it uses as much as digits as its factorization.
1300220132227 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2016, while the sum is 25.
Adding to 1300220132227 its reverse (7222310220031), we get a palindrome (8522530352258).
The spelling of 1300220132227 in words is "one trillion, three hundred billion, two hundred twenty million, one hundred thirty-two thousand, two hundred twenty-seven".
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