Base | Representation |
---|---|
bin | 1011110100110100011000… |
… | …0110011100100110011000 |
3 | 1201000222112200001222000021 |
4 | 2331031012012130212120 |
5 | 3201011140234200440 |
6 | 43353020005520224 |
7 | 2511236503226143 |
oct | 275150606344630 |
9 | 51028480058007 |
10 | 13002042100120 |
11 | 4163153891847 |
12 | 155ba71713074 |
13 | 734119a83198 |
14 | 32d43331d05a |
15 | 17832dab384a |
hex | bd34619c998 |
13002042100120 has 128 divisors (see below), whose sum is σ = 29737173392640. Its totient is φ = 5115621404160.
The previous prime is 13002042100057. The next prime is 13002042100123. The reversal of 13002042100120 is 2100124020031.
It is a junction number, because it is equal to n+sod(n) for n = 13002042100094 and 13002042100103.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13002042100123) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 796051474 + ... + 796067806.
It is an arithmetic number, because the mean of its divisors is an integer number (232321667130).
Almost surely, 213002042100120 is an apocalyptic number.
13002042100120 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 13002042100120, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (14868586696320).
13002042100120 is an abundant number, since it is smaller than the sum of its proper divisors (16735131292520).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13002042100120 is a wasteful number, since it uses less digits than its factorization.
13002042100120 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 17465 (or 17461 counting only the distinct ones).
The product of its (nonzero) digits is 96, while the sum is 16.
Adding to 13002042100120 its reverse (2100124020031), we get a palindrome (15102166120151).
The spelling of 13002042100120 in words is "thirteen trillion, two billion, forty-two million, one hundred thousand, one hundred twenty".
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