Base | Representation |
---|---|
bin | 1100101100011110000001… |
… | …0110100000000101110100 |
3 | 1211102101022212122122211102 |
4 | 3023013200112200011310 |
5 | 3312142203330132300 |
6 | 45404133553425232 |
7 | 2640304021460123 |
oct | 313074026400564 |
9 | 54371285578742 |
10 | 13958112739700 |
11 | 44a1669832a98 |
12 | 169521386b218 |
13 | 7a3323bab71c |
14 | 36380d391aba |
15 | 1931388c30d5 |
hex | cb1e05a0174 |
13958112739700 has 18 divisors (see below), whose sum is σ = 30289104645366. Its totient is φ = 5583245095840.
The previous prime is 13958112739693. The next prime is 13958112739793. The reversal of 13958112739700 is 793721185931.
It can be written as a sum of positive squares in 3 ways, for example, as 63155710864 + 13894957028836 = 251308^2 + 3727594^2 .
It is a super-2 number, since 2×139581127397002 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 69790563599 + ... + 69790563798.
Almost surely, 213958112739700 is an apocalyptic number.
13958112739700 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
13958112739700 is an abundant number, since it is smaller than the sum of its proper divisors (16330991905666).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13958112739700 is a wasteful number, since it uses less digits than its factorization.
13958112739700 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 139581127411 (or 139581127404 counting only the distinct ones).
The product of its (nonzero) digits is 2857680, while the sum is 56.
The spelling of 13958112739700 in words is "thirteen trillion, nine hundred fifty-eight billion, one hundred twelve million, seven hundred thirty-nine thousand, seven hundred".
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