Base | Representation |
---|---|
bin | 1011000000010100000011… |
… | …0110111000000101010111 |
3 | 1120211202020120102022102011 |
4 | 2300011000312320011113 |
5 | 3041221310304001101 |
6 | 41422400204525051 |
7 | 2356124360334304 |
oct | 260050066700527 |
9 | 46752216368364 |
10 | 12100011000151 |
11 | 3945649843538 |
12 | 1435091250787 |
13 | 69a045581199 |
14 | 2db9022a40ab |
15 | 15eb380ce751 |
hex | b0140db8157 |
12100011000151 has 2 divisors, whose sum is σ = 12100011000152. Its totient is φ = 12100011000150.
The previous prime is 12100011000013. The next prime is 12100011000373. The reversal of 12100011000151 is 15100011000121.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12100011000151 - 215 = 12100010967383 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12100011000851) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6050005500075 + 6050005500076.
It is an arithmetic number, because the mean of its divisors is an integer number (6050005500076).
Almost surely, 212100011000151 is an apocalyptic number.
12100011000151 is a deficient number, since it is larger than the sum of its proper divisors (1).
12100011000151 is an equidigital number, since it uses as much as digits as its factorization.
12100011000151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 10, while the sum is 13.
Adding to 12100011000151 its reverse (15100011000121), we get a palindrome (27200022000272).
The spelling of 12100011000151 in words is "twelve trillion, one hundred billion, eleven million, one hundred fifty-one", and thus it is an aban number.
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