Base | Representation |
---|---|
bin | 111100011001000101111101… |
… | …0000010101100000010100110 |
3 | 2120122212121001201102201011201 |
4 | 1320302023322002230002212 |
5 | 1024111403432032100200 |
6 | 5121444255431553114 |
7 | 216614653362242353 |
oct | 17062137202540246 |
9 | 2518777051381151 |
10 | 531214340112550 |
11 | 14429681129a7a6 |
12 | 4b6b4a9382b79a |
13 | 19a543b3300423 |
14 | 9526ca865d62a |
15 | 41631862c9c6a |
hex | 1e322fa0ac0a6 |
531214340112550 has 48 divisors (see below), whose sum is σ = 1023780922599600. Its totient is φ = 204827205744000.
The previous prime is 531214340112499. The next prime is 531214340112557. The reversal of 531214340112550 is 55211043412135.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (531214340112557) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 295027101 + ... + 296822200.
It is an arithmetic number, because the mean of its divisors is an integer number (21328769220825).
Almost surely, 2531214340112550 is an apocalyptic number.
531214340112550 is a gapful number since it is divisible by the number (50) formed by its first and last digit.
531214340112550 is a deficient number, since it is larger than the sum of its proper divisors (492566582487050).
531214340112550 is a wasteful number, since it uses less digits than its factorization.
531214340112550 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 591849961 (or 591849956 counting only the distinct ones).
The product of its (nonzero) digits is 72000, while the sum is 37.
Adding to 531214340112550 its reverse (55211043412135), we get a palindrome (586425383524685).
The spelling of 531214340112550 in words is "five hundred thirty-one trillion, two hundred fourteen billion, three hundred forty million, one hundred twelve thousand, five hundred fifty".
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