Base | Representation |
---|---|
bin | 101100011110101001001… |
… | …0110100111100100001011 |
3 | 210122102000212000012120212 |
4 | 1120331102112213210023 |
5 | 1300124132222420042 |
6 | 21000154551511335 |
7 | 1200441546460442 |
oct | 130752226474413 |
9 | 23572025005525 |
10 | 6113120123147 |
11 | 1a47617a39234 |
12 | 82891b68854b |
13 | 3546070ac1cb |
14 | 171c3b527d59 |
15 | a9039e51482 |
hex | 58f525a790b |
6113120123147 has 2 divisors, whose sum is σ = 6113120123148. Its totient is φ = 6113120123146.
The previous prime is 6113120123141. The next prime is 6113120123219. The reversal of 6113120123147 is 7413210213116.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 6113120123147 - 230 = 6112046381323 is a prime.
It is not a weakly prime, because it can be changed into another prime (6113120123141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3056560061573 + 3056560061574.
It is an arithmetic number, because the mean of its divisors is an integer number (3056560061574).
Almost surely, 26113120123147 is an apocalyptic number.
6113120123147 is a deficient number, since it is larger than the sum of its proper divisors (1).
6113120123147 is an equidigital number, since it uses as much as digits as its factorization.
6113120123147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048, while the sum is 32.
The spelling of 6113120123147 in words is "six trillion, one hundred thirteen billion, one hundred twenty million, one hundred twenty-three thousand, one hundred forty-seven".
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