Base | Representation |
---|---|
bin | 101110101110110001110… |
… | …010110010001100000001 |
3 | 102100222222122121211022202 |
4 | 232232301302302030001 |
5 | 410103242300400423 |
6 | 10455132302242545 |
7 | 451003405642142 |
oct | 56566162621401 |
9 | 12328878554282 |
10 | 3211323450113 |
11 | 1028a07406609 |
12 | 43a461886455 |
13 | 1a3a98a950cc |
14 | b1600c03ac9 |
15 | 58801d49728 |
hex | 2ebb1cb2301 |
3211323450113 has 2 divisors, whose sum is σ = 3211323450114. Its totient is φ = 3211323450112.
The previous prime is 3211323450109. The next prime is 3211323450119. The reversal of 3211323450113 is 3110543231123.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2612383666369 + 598939783744 = 1616287^2 + 773912^2 .
It is a cyclic number.
It is not a de Polignac number, because 3211323450113 - 22 = 3211323450109 is a prime.
It is not a weakly prime, because it can be changed into another prime (3211323450119) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1605661725056 + 1605661725057.
It is an arithmetic number, because the mean of its divisors is an integer number (1605661725057).
Almost surely, 23211323450113 is an apocalyptic number.
It is an amenable number.
3211323450113 is a deficient number, since it is larger than the sum of its proper divisors (1).
3211323450113 is an equidigital number, since it uses as much as digits as its factorization.
3211323450113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6480, while the sum is 29.
Adding to 3211323450113 its reverse (3110543231123), we get a palindrome (6321866681236).
The spelling of 3211323450113 in words is "three trillion, two hundred eleven billion, three hundred twenty-three million, four hundred fifty thousand, one hundred thirteen".
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