Base | Representation |
---|---|
bin | 1110000001110000101111… |
… | …0001010010111101010011 |
3 | 2000121110120000020200010212 |
4 | 3200130023301102331103 |
5 | 4010144133302011034 |
6 | 52445231301314335 |
7 | 3151206564446444 |
oct | 340341361227523 |
9 | 60543500220125 |
10 | 15423425032019 |
11 | 4a07046385174 |
12 | 18911b5a059ab |
13 | 87b5659b41c4 |
14 | 3b46d7ba52cb |
15 | 1bb2ea8282ce |
hex | e070bc52f53 |
15423425032019 has 2 divisors, whose sum is σ = 15423425032020. Its totient is φ = 15423425032018.
The previous prime is 15423425032013. The next prime is 15423425032079. The reversal of 15423425032019 is 91023052432451.
15423425032019 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 15423425032019 - 24 = 15423425032003 is a prime.
It is a super-2 number, since 2×154234250320192 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (15423425032013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7711712516009 + 7711712516010.
It is an arithmetic number, because the mean of its divisors is an integer number (7711712516010).
Almost surely, 215423425032019 is an apocalyptic number.
15423425032019 is a deficient number, since it is larger than the sum of its proper divisors (1).
15423425032019 is an equidigital number, since it uses as much as digits as its factorization.
15423425032019 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 259200, while the sum is 41.
The spelling of 15423425032019 in words is "fifteen trillion, four hundred twenty-three billion, four hundred twenty-five million, thirty-two thousand, nineteen".
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