Base | Representation |
---|---|
bin | 1100000000100001000110… |
… | …0111110000101000101111 |
3 | 1201202012022112001101111211 |
4 | 3000020101213300220233 |
5 | 3212304243411302112 |
6 | 44025215213055251 |
7 | 2531613161454016 |
oct | 300102147605057 |
9 | 51665275041454 |
10 | 13203025103407 |
11 | 423040a517957 |
12 | 1592a02487527 |
13 | 74a06993b0b9 |
14 | 33905bba5a7d |
15 | 17d6934ad0a7 |
hex | c02119f0a2f |
13203025103407 has 2 divisors, whose sum is σ = 13203025103408. Its totient is φ = 13203025103406.
The previous prime is 13203025103393. The next prime is 13203025103461. The reversal of 13203025103407 is 70430152030231.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13203025103407 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13203025103207) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6601512551703 + 6601512551704.
It is an arithmetic number, because the mean of its divisors is an integer number (6601512551704).
Almost surely, 213203025103407 is an apocalyptic number.
13203025103407 is a deficient number, since it is larger than the sum of its proper divisors (1).
13203025103407 is an equidigital number, since it uses as much as digits as its factorization.
13203025103407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 15120, while the sum is 31.
Adding to 13203025103407 its reverse (70430152030231), we get a palindrome (83633177133638).
The spelling of 13203025103407 in words is "thirteen trillion, two hundred three billion, twenty-five million, one hundred three thousand, four hundred seven".
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