Base | Representation |
---|---|
bin | 100110101111101101100… |
… | …1001010011101100010111 |
3 | 200212002010020111212221112 |
4 | 1031133123021103230113 |
5 | 1144221342112130012 |
6 | 15154200331205235 |
7 | 1056504650523566 |
oct | 115373311235427 |
9 | 20762106455845 |
10 | 5325141130007 |
11 | 177341a383956 |
12 | 72006aba681b |
13 | 2c820a740b5a |
14 | 145a498cbbdd |
15 | 937bc17e122 |
hex | 4d7db253b17 |
5325141130007 has 2 divisors, whose sum is σ = 5325141130008. Its totient is φ = 5325141130006.
The previous prime is 5325141129929. The next prime is 5325141130019. The reversal of 5325141130007 is 7000311415235.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5325141130007 - 212 = 5325141125911 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5325141130097) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2662570565003 + 2662570565004.
It is an arithmetic number, because the mean of its divisors is an integer number (2662570565004).
Almost surely, 25325141130007 is an apocalyptic number.
5325141130007 is a deficient number, since it is larger than the sum of its proper divisors (1).
5325141130007 is an equidigital number, since it uses as much as digits as its factorization.
5325141130007 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 12600, while the sum is 32.
The spelling of 5325141130007 in words is "five trillion, three hundred twenty-five billion, one hundred forty-one million, one hundred thirty thousand, seven".
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