Base | Representation |
---|---|
bin | 10111111010000111… |
… | …000110100000000101 |
3 | 2110021001112000200012 |
4 | 113322013012200011 |
5 | 410033234031432 |
6 | 15443150400005 |
7 | 1566102626525 |
oct | 277207064005 |
9 | 73231460605 |
10 | 25671002117 |
11 | a98367a932 |
12 | 4b85200005 |
13 | 2561561aa7 |
14 | 1357533685 |
15 | a03a67cb2 |
hex | 5fa1c6805 |
25671002117 has 2 divisors, whose sum is σ = 25671002118. Its totient is φ = 25671002116.
The previous prime is 25671002089. The next prime is 25671002141. The reversal of 25671002117 is 71120017652.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 25553301316 + 117700801 = 159854^2 + 10849^2 .
It is a cyclic number.
It is not a de Polignac number, because 25671002117 - 210 = 25671001093 is a prime.
It is a super-3 number, since 3×256710021173 (a number of 32 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (25671002417) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 12835501058 + 12835501059.
It is an arithmetic number, because the mean of its divisors is an integer number (12835501059).
Almost surely, 225671002117 is an apocalyptic number.
It is an amenable number.
25671002117 is a deficient number, since it is larger than the sum of its proper divisors (1).
25671002117 is an equidigital number, since it uses as much as digits as its factorization.
25671002117 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5880, while the sum is 32.
Adding to 25671002117 its reverse (71120017652), we get a palindrome (96791019769).
The spelling of 25671002117 in words is "twenty-five billion, six hundred seventy-one million, two thousand, one hundred seventeen".
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