Base | Representation |
---|---|
bin | 111110011111111010… |
… | …1110011001110010111 |
3 | 110211102200022012000111 |
4 | 1330333311303032113 |
5 | 4144333023213411 |
6 | 141354010223451 |
7 | 12460655065126 |
oct | 1747765631627 |
9 | 424380265014 |
10 | 134215054231 |
11 | 51a13998825 |
12 | 22018424b87 |
13 | c86c22176b |
14 | 66d31ddabd |
15 | 3757e18721 |
hex | 1f3fd73397 |
134215054231 has 2 divisors, whose sum is σ = 134215054232. Its totient is φ = 134215054230.
The previous prime is 134215054097. The next prime is 134215054271. The reversal of 134215054231 is 132450512431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 134215054231 - 215 = 134215021463 is a prime.
It is a super-2 number, since 2×1342150542312 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 134215054193 and 134215054202.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134215054271) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67107527115 + 67107527116.
It is an arithmetic number, because the mean of its divisors is an integer number (67107527116).
Almost surely, 2134215054231 is an apocalyptic number.
134215054231 is a deficient number, since it is larger than the sum of its proper divisors (1).
134215054231 is an equidigital number, since it uses as much as digits as its factorization.
134215054231 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 134215054231 its reverse (132450512431), we get a palindrome (266665566662).
The spelling of 134215054231 in words is "one hundred thirty-four billion, two hundred fifteen million, fifty-four thousand, two hundred thirty-one".
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