Base | Representation |
---|---|
bin | 10100100010010010001… |
… | …111001101010000111001 |
3 | 11222220120010220121021002 |
4 | 110202102033031100321 |
5 | 141110120344424423 |
6 | 3000144140054345 |
7 | 203645630212355 |
oct | 24422217152071 |
9 | 4886503817232 |
10 | 1411202733113 |
11 | 4a4540175508 |
12 | 1a9600b343b5 |
13 | a30ca834834 |
14 | 4c4343b0465 |
15 | 26a968e8228 |
hex | 148923cd439 |
1411202733113 has 2 divisors, whose sum is σ = 1411202733114. Its totient is φ = 1411202733112.
The previous prime is 1411202733107. The next prime is 1411202733181. The reversal of 1411202733113 is 3113372021141.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1201244496169 + 209958236944 = 1096013^2 + 458212^2 .
It is a cyclic number.
It is not a de Polignac number, because 1411202733113 - 236 = 1342483256377 is a prime.
It is a super-2 number, since 2×14112027331132 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1411202735113) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 705601366556 + 705601366557.
It is an arithmetic number, because the mean of its divisors is an integer number (705601366557).
Almost surely, 21411202733113 is an apocalyptic number.
It is an amenable number.
1411202733113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1411202733113 is an equidigital number, since it uses as much as digits as its factorization.
1411202733113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3024, while the sum is 29.
Adding to 1411202733113 its reverse (3113372021141), we get a palindrome (4524574754254).
The spelling of 1411202733113 in words is "one trillion, four hundred eleven billion, two hundred two million, seven hundred thirty-three thousand, one hundred thirteen".
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