Base | Representation |
---|---|
bin | 1010111011001100100110… |
… | …1011100100000110100111 |
3 | 1120112100100202022121202021 |
4 | 2232303021223210012213 |
5 | 3033301301210400403 |
6 | 41314142124324011 |
7 | 2346563224013161 |
oct | 256631153440647 |
9 | 46470322277667 |
10 | 12012112200103 |
11 | 3911343200034 |
12 | 14200400a3607 |
13 | 691977b79cb2 |
14 | 2d7564485931 |
15 | 15c6e14558bd |
hex | aecc9ae41a7 |
12012112200103 has 2 divisors, whose sum is σ = 12012112200104. Its totient is φ = 12012112200102.
The previous prime is 12012112200073. The next prime is 12012112200131. The reversal of 12012112200103 is 30100221121021.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-12012112200103 is a prime.
It is a super-2 number, since 2×120121122001032 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12012112200503) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6006056100051 + 6006056100052.
It is an arithmetic number, because the mean of its divisors is an integer number (6006056100052).
Almost surely, 212012112200103 is an apocalyptic number.
12012112200103 is a deficient number, since it is larger than the sum of its proper divisors (1).
12012112200103 is an equidigital number, since it uses as much as digits as its factorization.
12012112200103 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 48, while the sum is 16.
Adding to 12012112200103 its reverse (30100221121021), we get a palindrome (42112333321124).
The spelling of 12012112200103 in words is "twelve trillion, twelve billion, one hundred twelve million, two hundred thousand, one hundred three".
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