Base | Representation |
---|---|
bin | 10111011001010101… |
… | …100110000001110001 |
3 | 2101211201212210200011 |
4 | 113121111212001301 |
5 | 402422003400022 |
6 | 15312425254521 |
7 | 1546345164343 |
oct | 273125460161 |
9 | 71751783604 |
10 | 25121153137 |
11 | a721266193 |
12 | 4a51034441 |
13 | 24a4675272 |
14 | 13044c3293 |
15 | 9c0654977 |
hex | 5d9566071 |
25121153137 has 2 divisors, whose sum is σ = 25121153138. Its totient is φ = 25121153136.
The previous prime is 25121153117. The next prime is 25121153143. The reversal of 25121153137 is 73135112152.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 24478479936 + 642673201 = 156456^2 + 25351^2 .
It is a cyclic number.
It is not a de Polignac number, because 25121153137 - 215 = 25121120369 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 25121153099 and 25121153108.
It is not a weakly prime, because it can be changed into another prime (25121153117) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 12560576568 + 12560576569.
It is an arithmetic number, because the mean of its divisors is an integer number (12560576569).
Almost surely, 225121153137 is an apocalyptic number.
It is an amenable number.
25121153137 is a deficient number, since it is larger than the sum of its proper divisors (1).
25121153137 is an equidigital number, since it uses as much as digits as its factorization.
25121153137 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 6300, while the sum is 31.
Adding to 25121153137 its reverse (73135112152), we get a palindrome (98256265289).
The spelling of 25121153137 in words is "twenty-five billion, one hundred twenty-one million, one hundred fifty-three thousand, one hundred thirty-seven".
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