Base | Representation |
---|---|
bin | 100000100011110110101100… |
… | …101001101110001011100011 |
3 | 200210000220211212110120102201 |
4 | 200203312230221232023203 |
5 | 122232202432203443042 |
6 | 1224321452053554031 |
7 | 42106650134553004 |
oct | 4043665451561343 |
9 | 623026755416381 |
10 | 143201401234147 |
11 | 416a0411820073 |
12 | 14089483645917 |
13 | 61b9aac950ba9 |
14 | 2750db04191ab |
15 | 1184ee51e01b7 |
hex | 823daca6e2e3 |
143201401234147 has 2 divisors, whose sum is σ = 143201401234148. Its totient is φ = 143201401234146.
The previous prime is 143201401234127. The next prime is 143201401234163. The reversal of 143201401234147 is 741432104102341.
143201401234147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 143201401234147 - 219 = 143201400709859 is a prime.
It is not a weakly prime, because it can be changed into another prime (143201401234127) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71600700617073 + 71600700617074.
It is an arithmetic number, because the mean of its divisors is an integer number (71600700617074).
Almost surely, 2143201401234147 is an apocalyptic number.
143201401234147 is a deficient number, since it is larger than the sum of its proper divisors (1).
143201401234147 is an equidigital number, since it uses as much as digits as its factorization.
143201401234147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 64512, while the sum is 37.
Adding to 143201401234147 its reverse (741432104102341), we get a palindrome (884633505336488).
The spelling of 143201401234147 in words is "one hundred forty-three trillion, two hundred one billion, four hundred one million, two hundred thirty-four thousand, one hundred forty-seven".
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