Base | Representation |
---|---|
bin | 10100110111111001101… |
… | …100100100101111110111 |
3 | 12002010110201201122101222 |
4 | 110313321230210233313 |
5 | 142000134302234312 |
6 | 3014543235340555 |
7 | 205430046025406 |
oct | 24677154445767 |
9 | 5063421648358 |
10 | 1434413321207 |
11 | 503370a19919 |
12 | 1b1bba15175b |
13 | a5359417429 |
14 | 4d5d6c3963d |
15 | 274a4454672 |
hex | 14df9b24bf7 |
1434413321207 has 2 divisors, whose sum is σ = 1434413321208. Its totient is φ = 1434413321206.
The previous prime is 1434413321177. The next prime is 1434413321323. The reversal of 1434413321207 is 7021233144341.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1434413321207 - 210 = 1434413320183 is a prime.
It is a super-3 number, since 3×14344133212073 (a number of 37 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1434413321507) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 717206660603 + 717206660604.
It is an arithmetic number, because the mean of its divisors is an integer number (717206660604).
Almost surely, 21434413321207 is an apocalyptic number.
1434413321207 is a deficient number, since it is larger than the sum of its proper divisors (1).
1434413321207 is an equidigital number, since it uses as much as digits as its factorization.
1434413321207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 48384, while the sum is 35.
Adding to 1434413321207 its reverse (7021233144341), we get a palindrome (8455646465548).
The spelling of 1434413321207 in words is "one trillion, four hundred thirty-four billion, four hundred thirteen million, three hundred twenty-one thousand, two hundred seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.104 sec. • engine limits •