Base | Representation |
---|---|
bin | 10010101110010101… |
… | …00111001000100001 |
3 | 221221120020111000212 |
4 | 21113022213020201 |
5 | 131041344142423 |
6 | 4341252223505 |
7 | 504051300353 |
oct | 112712471041 |
9 | 27846214025 |
10 | 10052334113 |
11 | 429930a448 |
12 | 1b4660ab95 |
13 | c427b20c3 |
14 | 6b50a70d3 |
15 | 3dc799578 |
hex | 2572a7221 |
10052334113 has 2 divisors, whose sum is σ = 10052334114. Its totient is φ = 10052334112.
The previous prime is 10052334109. The next prime is 10052334137. The reversal of 10052334113 is 31143325001.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 5900851489 + 4151482624 = 76817^2 + 64432^2 .
It is an emirp because it is prime and its reverse (31143325001) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 10052334113 - 22 = 10052334109 is a prime.
It is a super-2 number, since 2×100523341132 (a number of 21 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (10052334163) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5026167056 + 5026167057.
It is an arithmetic number, because the mean of its divisors is an integer number (5026167057).
Almost surely, 210052334113 is an apocalyptic number.
It is an amenable number.
10052334113 is a deficient number, since it is larger than the sum of its proper divisors (1).
10052334113 is an equidigital number, since it uses as much as digits as its factorization.
10052334113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1080, while the sum is 23.
Adding to 10052334113 its reverse (31143325001), we get a palindrome (41195659114).
The spelling of 10052334113 in words is "ten billion, fifty-two million, three hundred thirty-four thousand, one hundred thirteen".
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