Base | Representation |
---|---|
bin | 10011111011101001010… |
… | …101101101100100011011 |
3 | 11211221110202021001102012 |
4 | 103323221111231210123 |
5 | 134420133340230011 |
6 | 2525123232500135 |
7 | 200646541104653 |
oct | 23735125554433 |
9 | 4757422231365 |
10 | 1369714383131 |
11 | 488990074066 |
12 | 1a15626aa64b |
13 | 9c21830b913 |
14 | 4a41a293363 |
15 | 25969418a8b |
hex | 13ee956d91b |
1369714383131 has 2 divisors, whose sum is σ = 1369714383132. Its totient is φ = 1369714383130.
The previous prime is 1369714383119. The next prime is 1369714383139. The reversal of 1369714383131 is 1313834179631.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1369714383131 - 26 = 1369714383067 is a prime.
It is a super-2 number, since 2×13697143831312 (a number of 25 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 1369714383131.
It is not a weakly prime, because it can be changed into another prime (1369714383139) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 684857191565 + 684857191566.
It is an arithmetic number, because the mean of its divisors is an integer number (684857191566).
Almost surely, 21369714383131 is an apocalyptic number.
1369714383131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1369714383131 is an equidigital number, since it uses as much as digits as its factorization.
1369714383131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 979776, while the sum is 50.
The spelling of 1369714383131 in words is "one trillion, three hundred sixty-nine billion, seven hundred fourteen million, three hundred eighty-three thousand, one hundred thirty-one".
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