Base | Representation |
---|---|
bin | 11101000111111111010… |
… | …011101001001111110111 |
3 | 21002100002000100120110111 |
4 | 131013333103221033313 |
5 | 230242423420214112 |
6 | 4131241132203451 |
7 | 264412424633125 |
oct | 35077723511767 |
9 | 7070060316414 |
10 | 2001443132407 |
11 | 70189751a5a4 |
12 | 283a87309587 |
13 | 116973535384 |
14 | 6cc283cc915 |
15 | 370dea425a7 |
hex | 1d1ff4e93f7 |
2001443132407 has 2 divisors, whose sum is σ = 2001443132408. Its totient is φ = 2001443132406.
The previous prime is 2001443132401. The next prime is 2001443132459. The reversal of 2001443132407 is 7042313441002.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2001443132407 - 231 = 1999295648759 is a prime.
It is a super-2 number, since 2×20014431324072 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2001443132401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1000721566203 + 1000721566204.
It is an arithmetic number, because the mean of its divisors is an integer number (1000721566204).
Almost surely, 22001443132407 is an apocalyptic number.
2001443132407 is a deficient number, since it is larger than the sum of its proper divisors (1).
2001443132407 is an equidigital number, since it uses as much as digits as its factorization.
2001443132407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 16128, while the sum is 31.
Adding to 2001443132407 its reverse (7042313441002), we get a palindrome (9043756573409).
The spelling of 2001443132407 in words is "two trillion, one billion, four hundred forty-three million, one hundred thirty-two thousand, four hundred seven".
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