Base | Representation |
---|---|
bin | 10001011110110011101… |
… | …111110101000001100001 |
3 | 11020211210110121000001021 |
4 | 101132303233311001201 |
5 | 124140241223111301 |
6 | 2315512551205441 |
7 | 152535422365561 |
oct | 21366357650141 |
9 | 4224713530037 |
10 | 1201311535201 |
11 | 423523052988 |
12 | 1749a4976881 |
13 | 8938b23123b |
14 | 422027acda1 |
15 | 213aee088a1 |
hex | 117b3bf5061 |
1201311535201 has 2 divisors, whose sum is σ = 1201311535202. Its totient is φ = 1201311535200.
The previous prime is 1201311535187. The next prime is 1201311535283. The reversal of 1201311535201 is 1025351131021.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 801383040000 + 399928495201 = 895200^2 + 632399^2 .
It is a cyclic number.
It is not a de Polignac number, because 1201311535201 - 211 = 1201311533153 is a prime.
It is a super-2 number, since 2×12013115352012 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1201311535301) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 600655767600 + 600655767601.
It is an arithmetic number, because the mean of its divisors is an integer number (600655767601).
Almost surely, 21201311535201 is an apocalyptic number.
It is an amenable number.
1201311535201 is a deficient number, since it is larger than the sum of its proper divisors (1).
1201311535201 is an equidigital number, since it uses as much as digits as its factorization.
1201311535201 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 900, while the sum is 25.
Adding to 1201311535201 its reverse (1025351131021), we get a palindrome (2226662666222).
The spelling of 1201311535201 in words is "one trillion, two hundred one billion, three hundred eleven million, five hundred thirty-five thousand, two hundred one".
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