Base | Representation |
---|---|
bin | 1010001000001010000010… |
… | …1100010110100010010001 |
3 | 1110102112000210200011011121 |
4 | 2202002200230112202101 |
5 | 2424414443302244213 |
6 | 35403244102242241 |
7 | 2226331622420035 |
oct | 242024054264221 |
9 | 43375023604147 |
10 | 11135251212433 |
11 | 3603485017727 |
12 | 12ba105126381 |
13 | 62a08520297a |
14 | 2a6d402161c5 |
15 | 1449c054788d |
hex | a20a0b16891 |
11135251212433 has 2 divisors, whose sum is σ = 11135251212434. Its totient is φ = 11135251212432.
The previous prime is 11135251212413. The next prime is 11135251212509. The reversal of 11135251212433 is 33421215253111.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 9521495118864 + 1613756093569 = 3085692^2 + 1270337^2 .
It is a cyclic number.
It is not a de Polignac number, because 11135251212433 - 25 = 11135251212401 is a prime.
It is not a weakly prime, because it can be changed into another prime (11135251212413) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5567625606216 + 5567625606217.
It is an arithmetic number, because the mean of its divisors is an integer number (5567625606217).
Almost surely, 211135251212433 is an apocalyptic number.
It is an amenable number.
11135251212433 is a deficient number, since it is larger than the sum of its proper divisors (1).
11135251212433 is an equidigital number, since it uses as much as digits as its factorization.
11135251212433 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 21600, while the sum is 34.
Adding to 11135251212433 its reverse (33421215253111), we get a palindrome (44556466465544).
The spelling of 11135251212433 in words is "eleven trillion, one hundred thirty-five billion, two hundred fifty-one million, two hundred twelve thousand, four hundred thirty-three".
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