Base | Representation |
---|---|
bin | 1010000010001001011100… |
… | …1111000101111010110011 |
3 | 1110001122112222101110011211 |
4 | 2200202113033011322303 |
5 | 2421222030403112042 |
6 | 35244011552220551 |
7 | 2216015401052233 |
oct | 240422717057263 |
9 | 43048488343154 |
10 | 11032013332147 |
11 | 3573717a16205 |
12 | 12a20b2b3a757 |
13 | 620411950942 |
14 | 2a1d49149cc3 |
15 | 141e7be2b017 |
hex | a08973c5eb3 |
11032013332147 has 2 divisors, whose sum is σ = 11032013332148. Its totient is φ = 11032013332146.
The previous prime is 11032013332129. The next prime is 11032013332159. The reversal of 11032013332147 is 74123331023011.
11032013332147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11032013332147 - 219 = 11032012807859 is a prime.
It is not a weakly prime, because it can be changed into another prime (11032013332127) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5516006666073 + 5516006666074.
It is an arithmetic number, because the mean of its divisors is an integer number (5516006666074).
Almost surely, 211032013332147 is an apocalyptic number.
11032013332147 is a deficient number, since it is larger than the sum of its proper divisors (1).
11032013332147 is an equidigital number, since it uses as much as digits as its factorization.
11032013332147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 9072, while the sum is 31.
Adding to 11032013332147 its reverse (74123331023011), we get a palindrome (85155344355158).
The spelling of 11032013332147 in words is "eleven trillion, thirty-two billion, thirteen million, three hundred thirty-two thousand, one hundred forty-seven".
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