Base | Representation |
---|---|
bin | 11101100100001110110010… |
… | …100001011111010111110111 |
3 | 122001102001020101222011211001 |
4 | 131210032302201133113313 |
5 | 114020430322240310403 |
6 | 1140320221213024131 |
7 | 36250404116145265 |
oct | 3544166241372767 |
9 | 561361211864731 |
10 | 130033130010103 |
11 | 38483816a30171 |
12 | 12701360185047 |
13 | 57730c1251a68 |
14 | 24178cc1b3435 |
15 | 10076d7d97c1d |
hex | 7643b285f5f7 |
130033130010103 has 2 divisors, whose sum is σ = 130033130010104. Its totient is φ = 130033130010102.
The previous prime is 130033130010089. The next prime is 130033130010107. The reversal of 130033130010103 is 301010031330031.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130033130010103 - 25 = 130033130010071 is a prime.
It is a super-2 number, since 2×1300331300101032 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130033130010107) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65016565005051 + 65016565005052.
It is an arithmetic number, because the mean of its divisors is an integer number (65016565005052).
Almost surely, 2130033130010103 is an apocalyptic number.
130033130010103 is a deficient number, since it is larger than the sum of its proper divisors (1).
130033130010103 is an equidigital number, since it uses as much as digits as its factorization.
130033130010103 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 243, while the sum is 19.
Adding to 130033130010103 its reverse (301010031330031), we get a palindrome (431043161340134).
The spelling of 130033130010103 in words is "one hundred thirty trillion, thirty-three billion, one hundred thirty million, ten thousand, one hundred three".
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