Base | Representation |
---|---|
bin | 10011010010000111011… |
… | …001011000001101111111 |
3 | 11200200100222220002100221 |
4 | 103102013121120031333 |
5 | 133202322101113002 |
6 | 2452430304402211 |
7 | 164510513640241 |
oct | 23220731301577 |
9 | 4620328802327 |
10 | 1325121504127 |
11 | 470a875a8053 |
12 | 194998616367 |
13 | 97c5c863395 |
14 | 481c9b20091 |
15 | 247095e8c37 |
hex | 1348765837f |
1325121504127 has 2 divisors, whose sum is σ = 1325121504128. Its totient is φ = 1325121504126.
The previous prime is 1325121504019. The next prime is 1325121504173. The reversal of 1325121504127 is 7214051215231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1325121504127 - 231 = 1322974020479 is a prime.
It is a super-3 number, since 3×13251215041273 (a number of 37 digits) contains 333 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1325121504092 and 1325121504101.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1325121534127) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 662560752063 + 662560752064.
It is an arithmetic number, because the mean of its divisors is an integer number (662560752064).
Almost surely, 21325121504127 is an apocalyptic number.
1325121504127 is a deficient number, since it is larger than the sum of its proper divisors (1).
1325121504127 is an equidigital number, since it uses as much as digits as its factorization.
1325121504127 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 16800, while the sum is 34.
Adding to 1325121504127 its reverse (7214051215231), we get a palindrome (8539172719358).
The spelling of 1325121504127 in words is "one trillion, three hundred twenty-five billion, one hundred twenty-one million, five hundred four thousand, one hundred twenty-seven".
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