Base | Representation |
---|---|
bin | 1110001100000100110011… |
… | …10000101000010010000111 |
3 | 11002110210210210222222122100 |
4 | 13012002121300220102013 |
5 | 13042200100324200403 |
6 | 150205351001403143 |
7 | 6400133240644266 |
oct | 706023160502207 |
9 | 132423723888570 |
10 | 31201222100103 |
11 | 9a3a400509030 |
12 | 35bb00a1994b3 |
13 | 1454354c886c3 |
14 | 79c20bd185dd |
15 | 391936e182a3 |
hex | 1c6099c28487 |
31201222100103 has 48 divisors (see below), whose sum is σ = 49239486691200. Its totient is φ = 18881406815040.
The previous prime is 31201222100089. The next prime is 31201222100117. The reversal of 31201222100103 is 30100122210213.
It is an interprime number because it is at equal distance from previous prime (31201222100089) and next prime (31201222100117).
It is not a de Polignac number, because 31201222100103 - 25 = 31201222100071 is a prime.
It is a super-3 number, since 3×312012221001033 (a number of 41 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (31201222100143) by changing a digit.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 552117813 + ... + 552174321.
It is an arithmetic number, because the mean of its divisors is an integer number (1025822639400).
Almost surely, 231201222100103 is an apocalyptic number.
31201222100103 is a gapful number since it is divisible by the number (33) formed by its first and last digit.
31201222100103 is a deficient number, since it is larger than the sum of its proper divisors (18038264591097).
31201222100103 is a wasteful number, since it uses less digits than its factorization.
31201222100103 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 64812 (or 64809 counting only the distinct ones).
The product of its (nonzero) digits is 144, while the sum is 18.
Adding to 31201222100103 its reverse (30100122210213), we get a palindrome (61301344310316).
It can be divided in two parts, 3120122 and 2100103, that added together give a palindrome (5220225).
The spelling of 31201222100103 in words is "thirty-one trillion, two hundred one billion, two hundred twenty-two million, one hundred thousand, one hundred three".
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