Base | Representation |
---|---|
bin | 1110001001100000110001… |
… | …11111101110010010110111 |
3 | 11002011101111210020021022211 |
4 | 13010300120333232102313 |
5 | 13034224224041400403 |
6 | 150101104533434251 |
7 | 6360565112163646 |
oct | 704603077562267 |
9 | 132141453207284 |
10 | 31113162450103 |
11 | 9a060220a2688 |
12 | 35a5b3319b987 |
13 | 1448c60156c26 |
14 | 797c569c935d |
15 | 38e4d10ba06d |
hex | 1c4c18fee4b7 |
31113162450103 has 2 divisors, whose sum is σ = 31113162450104. Its totient is φ = 31113162450102.
The previous prime is 31113162450091. The next prime is 31113162450113. The reversal of 31113162450103 is 30105426131113.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 31113162450103 - 221 = 31113160352951 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (31113162450113) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15556581225051 + 15556581225052.
It is an arithmetic number, because the mean of its divisors is an integer number (15556581225052).
Almost surely, 231113162450103 is an apocalyptic number.
31113162450103 is a deficient number, since it is larger than the sum of its proper divisors (1).
31113162450103 is an equidigital number, since it uses as much as digits as its factorization.
31113162450103 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6480, while the sum is 31.
Adding to 31113162450103 its reverse (30105426131113), we get a palindrome (61218588581216).
The spelling of 31113162450103 in words is "thirty-one trillion, one hundred thirteen billion, one hundred sixty-two million, four hundred fifty thousand, one hundred three".
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