Base | Representation |
---|---|
bin | 1001000111011010000… |
… | …01011110001110101111 |
3 | 1002221110102211112112112 |
4 | 10203231001132032233 |
5 | 20112430320401341 |
6 | 355515535554235 |
7 | 31425512016461 |
oct | 4435501361657 |
9 | 1087412745475 |
10 | 313214231471 |
11 | 1109192a4a59 |
12 | 5085298b97b |
13 | 236c76cb49b |
14 | 11234116131 |
15 | 823282baeb |
hex | 48ed05e3af |
313214231471 has 2 divisors, whose sum is σ = 313214231472. Its totient is φ = 313214231470.
The previous prime is 313214231419. The next prime is 313214231509. The reversal of 313214231471 is 174132412313.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 313214231471 - 210 = 313214230447 is a prime.
It is a super-3 number, since 3×3132142314713 (a number of 35 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (313214231411) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 156607115735 + 156607115736.
It is an arithmetic number, because the mean of its divisors is an integer number (156607115736).
Almost surely, 2313214231471 is an apocalyptic number.
313214231471 is a deficient number, since it is larger than the sum of its proper divisors (1).
313214231471 is an equidigital number, since it uses as much as digits as its factorization.
313214231471 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 12096, while the sum is 32.
Adding to 313214231471 its reverse (174132412313), we get a palindrome (487346643784).
It can be divided in two parts, 3132142 and 31471, that added together give a palindrome (3163613).
The spelling of 313214231471 in words is "three hundred thirteen billion, two hundred fourteen million, two hundred thirty-one thousand, four hundred seventy-one".
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