Base | Representation |
---|---|
bin | 1100010011010001100… |
… | …1100011100111001111 |
3 | 202012111010101221201202 |
4 | 3010310121203213033 |
5 | 11430302110342312 |
6 | 241030153100115 |
7 | 21161005500125 |
oct | 3046431434717 |
9 | 665433357652 |
10 | 211332512207 |
11 | 8169776a373 |
12 | 34b5a9b763b |
13 | 16c0c0b8bbb |
14 | a32b1c9915 |
15 | 576d2e59c2 |
hex | 31346639cf |
211332512207 has 2 divisors, whose sum is σ = 211332512208. Its totient is φ = 211332512206.
The previous prime is 211332512173. The next prime is 211332512243. The reversal of 211332512207 is 702215233112.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 211332512207 - 216 = 211332446671 is a prime.
It is a super-2 number, since 2×2113325122072 (a number of 23 digits) contains 22 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 211332512207.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (211332512297) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 105666256103 + 105666256104.
It is an arithmetic number, because the mean of its divisors is an integer number (105666256104).
Almost surely, 2211332512207 is an apocalyptic number.
211332512207 is a deficient number, since it is larger than the sum of its proper divisors (1).
211332512207 is an equidigital number, since it uses as much as digits as its factorization.
211332512207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5040, while the sum is 29.
Adding to 211332512207 its reverse (702215233112), we get a palindrome (913547745319).
The spelling of 211332512207 in words is "two hundred eleven billion, three hundred thirty-two million, five hundred twelve thousand, two hundred seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.074 sec. • engine limits •