Base | Representation |
---|---|
bin | 11011011000100010000111… |
… | …010000001000000010110011 |
3 | 120210102022000102221011111021 |
4 | 123120202013100020002303 |
5 | 111241134043421200001 |
6 | 1104050120514200311 |
7 | 34240002531434101 |
oct | 3330420720100263 |
9 | 523368012834437 |
10 | 120433152131251 |
11 | 3541245a768493 |
12 | 11610902b83097 |
13 | 5227a4a1ab60b |
14 | 21a4dd2164c71 |
15 | ddcb2176cba1 |
hex | 6d88874080b3 |
120433152131251 has 2 divisors, whose sum is σ = 120433152131252. Its totient is φ = 120433152131250.
The previous prime is 120433152131249. The next prime is 120433152131371. The reversal of 120433152131251 is 152131251334021.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 120433152131251 - 21 = 120433152131249 is a prime.
Together with 120433152131249, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (120433152131651) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60216576065625 + 60216576065626.
It is an arithmetic number, because the mean of its divisors is an integer number (60216576065626).
Almost surely, 2120433152131251 is an apocalyptic number.
120433152131251 is a deficient number, since it is larger than the sum of its proper divisors (1).
120433152131251 is an equidigital number, since it uses as much as digits as its factorization.
120433152131251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 21600, while the sum is 34.
The spelling of 120433152131251 in words is "one hundred twenty trillion, four hundred thirty-three billion, one hundred fifty-two million, one hundred thirty-one thousand, two hundred fifty-one".
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