Base | Representation |
---|---|
bin | 1011111011101100101110… |
… | …0100110101011101101000 |
3 | 1201110021122110201112011111 |
4 | 2332323023210311131220 |
5 | 3204430230330144000 |
6 | 43523205124040104 |
7 | 2522622626054635 |
oct | 276731344653550 |
9 | 51407573645144 |
10 | 13120245553000 |
11 | 41a92a0630483 |
12 | 157a95b833034 |
13 | 7423069b9001 |
14 | 335047c0b18c |
15 | 17b44ae93cba |
hex | beecb935768 |
13120245553000 has 32 divisors (see below), whose sum is σ = 30701374596360. Its totient is φ = 5248098220800.
The previous prime is 13120245552959. The next prime is 13120245553153. The reversal of 13120245553000 is 35554202131.
It can be written as a sum of positive squares in 4 ways, for example, as 774565448836 + 12345680104164 = 880094^2 + 3513642^2 .
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 6560121777 + ... + 6560123776.
Almost surely, 213120245553000 is an apocalyptic number.
13120245553000 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
13120245553000 is an abundant number, since it is smaller than the sum of its proper divisors (17581129043360).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13120245553000 is a wasteful number, since it uses less digits than its factorization.
13120245553000 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 13120245574 (or 13120245560 counting only the distinct ones).
The product of its (nonzero) digits is 18000, while the sum is 31.
Adding to 13120245553000 its reverse (35554202131), we get a palindrome (13155799755131).
The spelling of 13120245553000 in words is "thirteen trillion, one hundred twenty billion, two hundred forty-five million, five hundred fifty-three thousand".
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