Base | Representation |
---|---|
bin | 11001101101111000011010… |
… | …011111011001110000110111 |
3 | 112211110122111010012020120112 |
4 | 121231320122133121300313 |
5 | 104311044210440244341 |
6 | 1040315151242453235 |
7 | 32552334235500353 |
oct | 3155703237316067 |
9 | 484418433166515 |
10 | 113104113212471 |
11 | 33047204321417 |
12 | 1082840280421b |
13 | 4b15897820b14 |
14 | 1dd03a827c463 |
15 | d1216e32aaeb |
hex | 66de1a7d9c37 |
113104113212471 has 2 divisors, whose sum is σ = 113104113212472. Its totient is φ = 113104113212470.
The previous prime is 113104113212459. The next prime is 113104113212507. The reversal of 113104113212471 is 174212311401311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 113104113212471 - 218 = 113104112950327 is a prime.
It is a super-3 number, since 3×1131041132124713 (a number of 43 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113104113202471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56552056606235 + 56552056606236.
It is an arithmetic number, because the mean of its divisors is an integer number (56552056606236).
Almost surely, 2113104113212471 is an apocalyptic number.
113104113212471 is a deficient number, since it is larger than the sum of its proper divisors (1).
113104113212471 is an equidigital number, since it uses as much as digits as its factorization.
113104113212471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4032, while the sum is 32.
Adding to 113104113212471 its reverse (174212311401311), we get a palindrome (287316424613782).
The spelling of 113104113212471 in words is "one hundred thirteen trillion, one hundred four billion, one hundred thirteen million, two hundred twelve thousand, four hundred seventy-one".
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