Base | Representation |
---|---|
bin | 111110110010000100001… |
… | …001100001001111001011 |
3 | 120021110010110111101121111 |
4 | 332302010021201033023 |
5 | 1031141304222100011 |
6 | 13101554102551151 |
7 | 623462653033462 |
oct | 76620411411713 |
9 | 16243113441544 |
10 | 4314364253131 |
11 | 1413791166949 |
12 | 59819ba384b7 |
13 | 253ac583caba |
14 | 10cb600959d9 |
15 | 7735e74e121 |
hex | 3ec842613cb |
4314364253131 has 2 divisors, whose sum is σ = 4314364253132. Its totient is φ = 4314364253130.
The previous prime is 4314364253117. The next prime is 4314364253137. The reversal of 4314364253131 is 1313524634134.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 4314364253131 - 25 = 4314364253099 is a prime.
It is a super-2 number, since 2×43143642531312 (a number of 26 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (4314364253137) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2157182126565 + 2157182126566.
It is an arithmetic number, because the mean of its divisors is an integer number (2157182126566).
Almost surely, 24314364253131 is an apocalyptic number.
4314364253131 is a deficient number, since it is larger than the sum of its proper divisors (1).
4314364253131 is an equidigital number, since it uses as much as digits as its factorization.
4314364253131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 311040, while the sum is 40.
Adding to 4314364253131 its reverse (1313524634134), we get a palindrome (5627888887265).
The spelling of 4314364253131 in words is "four trillion, three hundred fourteen billion, three hundred sixty-four million, two hundred fifty-three thousand, one hundred thirty-one".
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