Base | Representation |
---|---|
bin | 111010010111101001011001… |
… | …1011101001111100010000100 |
3 | 2111022212210112121102100202122 |
4 | 1310233102303131033202010 |
5 | 1014243412111221220200 |
6 | 5015543242054425112 |
7 | 213100421340020306 |
oct | 16457226335174204 |
9 | 2438783477370678 |
10 | 513423401351300 |
11 | 139657707162510 |
12 | 49700a8a362198 |
13 | 19063807c41c6b |
14 | 90adb7a569976 |
15 | 3e554c311c385 |
hex | 1d2f4b374f884 |
513423401351300 has 72 divisors (see below), whose sum is σ = 1215418181382720. Its totient is φ = 186698655875200.
The previous prime is 513423401351297. The next prime is 513423401351321. The reversal of 513423401351300 is 3153104324315.
It is an unprimeable number.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 316385582 + ... + 318004218.
It is an arithmetic number, because the mean of its divisors is an integer number (16880808074760).
Almost surely, 2513423401351300 is an apocalyptic number.
513423401351300 is a gapful number since it is divisible by the number (50) formed by its first and last digit.
It is an amenable number.
513423401351300 is an abundant number, since it is smaller than the sum of its proper divisors (701994780031420).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
513423401351300 is a wasteful number, since it uses less digits than its factorization.
513423401351300 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1907021 (or 1907014 counting only the distinct ones).
The product of its (nonzero) digits is 64800, while the sum is 35.
Adding to 513423401351300 its reverse (3153104324315), we get a palindrome (516576505675615).
The spelling of 513423401351300 in words is "five hundred thirteen trillion, four hundred twenty-three billion, four hundred one million, three hundred fifty-one thousand, three hundred".
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