Base | Representation |
---|---|
bin | 11101010101111100101… |
… | …011111000100010000001 |
3 | 21010202202120200120211201 |
4 | 131111330223320202001 |
5 | 231014122433242403 |
6 | 4142200313541201 |
7 | 265453024202356 |
oct | 35257453704201 |
9 | 7122676616751 |
10 | 2016431540353 |
11 | 7081890442aa |
12 | 28696aa14801 |
13 | 1181c1853009 |
14 | 6d84ac7662d |
15 | 376ba81721d |
hex | 1d57caf8881 |
2016431540353 has 2 divisors, whose sum is σ = 2016431540354. Its totient is φ = 2016431540352.
The previous prime is 2016431540311. The next prime is 2016431540359. The reversal of 2016431540353 is 3530451346102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2008421827344 + 8009713009 = 1417188^2 + 89497^2 .
It is a cyclic number.
It is not a de Polignac number, because 2016431540353 - 213 = 2016431532161 is a prime.
It is not a weakly prime, because it can be changed into another prime (2016431540359) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1008215770176 + 1008215770177.
It is an arithmetic number, because the mean of its divisors is an integer number (1008215770177).
Almost surely, 22016431540353 is an apocalyptic number.
It is an amenable number.
2016431540353 is a deficient number, since it is larger than the sum of its proper divisors (1).
2016431540353 is an equidigital number, since it uses as much as digits as its factorization.
2016431540353 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 129600, while the sum is 37.
Adding to 2016431540353 its reverse (3530451346102), we get a palindrome (5546882886455).
The spelling of 2016431540353 in words is "two trillion, sixteen billion, four hundred thirty-one million, five hundred forty thousand, three hundred fifty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.069 sec. • engine limits •