Base | Representation |
---|---|
bin | 11011010100001000000011… |
… | …111011001010101011011011 |
3 | 120202100100022020022022012122 |
4 | 123110020003323022223123 |
5 | 111221203324041023003 |
6 | 1103255033133531455 |
7 | 34206061564306151 |
oct | 3324100373125333 |
9 | 522310266268178 |
10 | 120130301111003 |
11 | 35305a7a206507 |
12 | 11582082943b8b |
13 | 52053147caba1 |
14 | 21944a251b3d1 |
15 | dd4ce8b47238 |
hex | 6d4203ecaadb |
120130301111003 has 2 divisors, whose sum is σ = 120130301111004. Its totient is φ = 120130301111002.
The previous prime is 120130301110997. The next prime is 120130301111057. The reversal of 120130301111003 is 300111103031021.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 120130301111003 - 28 = 120130301110747 is a prime.
It is a super-2 number, since 2×1201303011110032 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (120130301111063) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60065150555501 + 60065150555502.
It is an arithmetic number, because the mean of its divisors is an integer number (60065150555502).
Almost surely, 2120130301111003 is an apocalyptic number.
120130301111003 is a deficient number, since it is larger than the sum of its proper divisors (1).
120130301111003 is an equidigital number, since it uses as much as digits as its factorization.
120130301111003 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54, while the sum is 17.
Adding to 120130301111003 its reverse (300111103031021), we get a palindrome (420241404142024).
The spelling of 120130301111003 in words is "one hundred twenty trillion, one hundred thirty billion, three hundred one million, one hundred eleven thousand, three".
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