Base | Representation |
---|---|
bin | 11101000111110010000… |
… | …000011101100100110011 |
3 | 21002022111110111021102011 |
4 | 131013302000131210303 |
5 | 230241444310000201 |
6 | 4131203041503351 |
7 | 264404045222356 |
oct | 35076200354463 |
9 | 7068443437364 |
10 | 2001220000051 |
11 | 701792577998 |
12 | 283a24641b57 |
13 | 11693923bcac |
14 | 6cc0690829d |
15 | 370ca169151 |
hex | 1d1f201d933 |
2001220000051 has 2 divisors, whose sum is σ = 2001220000052. Its totient is φ = 2001220000050.
The previous prime is 2001219999979. The next prime is 2001220000123. The reversal of 2001220000051 is 1500000221002.
It is a balanced prime because it is at equal distance from previous prime (2001219999979) and next prime (2001220000123).
It is a cyclic number.
It is not a de Polignac number, because 2001220000051 - 235 = 1966860261683 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 2001219999986 and 2001220000040.
It is not a weakly prime, because it can be changed into another prime (2001220000651) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1000610000025 + 1000610000026.
It is an arithmetic number, because the mean of its divisors is an integer number (1000610000026).
Almost surely, 22001220000051 is an apocalyptic number.
2001220000051 is a deficient number, since it is larger than the sum of its proper divisors (1).
2001220000051 is an equidigital number, since it uses as much as digits as its factorization.
2001220000051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40, while the sum is 13.
Adding to 2001220000051 its reverse (1500000221002), we get a palindrome (3501220221053).
The spelling of 2001220000051 in words is "two trillion, one billion, two hundred twenty million, fifty-one", and thus it is an aban number.
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