Base | Representation |
---|---|
bin | 11100000010101101001110… |
… | …111011001000111001000111 |
3 | 121011200100200001022200101111 |
4 | 130002231032323020321013 |
5 | 112131130010331433112 |
6 | 1114145342140105451 |
7 | 34656251556220243 |
oct | 3402551673107107 |
9 | 534610601280344 |
10 | 123331310030407 |
11 | 3632a5732a9476 |
12 | 119ba526235287 |
13 | 53a812868ab8b |
14 | 22653a54c5423 |
15 | e3d1e51a45a7 |
hex | 702b4eec8e47 |
123331310030407 has 2 divisors, whose sum is σ = 123331310030408. Its totient is φ = 123331310030406.
The previous prime is 123331310030387. The next prime is 123331310030419. The reversal of 123331310030407 is 704030013133321.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-123331310030407 is a prime.
It is a super-3 number, since 3×1233313100304073 (a number of 43 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 a weakly prime, because it can be changed into another prime (123331310034407) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 61665655015203 + 61665655015204.
It is an arithmetic number, because the mean of its divisors is an integer number (61665655015204).
Almost surely, 2123331310030407 is an apocalyptic number.
123331310030407 is a deficient number, since it is larger than the sum of its proper divisors (1).
123331310030407 is an equidigital number, since it uses as much as digits as its factorization.
123331310030407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 13608, while the sum is 31.
Adding to 123331310030407 its reverse (704030013133321), we get a palindrome (827361323163728).
The spelling of 123331310030407 in words is "one hundred twenty-three trillion, three hundred thirty-one billion, three hundred ten million, thirty thousand, four hundred seven".
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