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260113140113 = 1699153097787
BaseRepresentation
bin1111001000111111110…
…0111000100110010001
3220212101201202121022012
43302033330320212101
513230202430440423
6315254435152305
724535563132161
oct3621774704621
9825351677265
10260113140113
11a034a0a5092
12424b3445695
131b6b4340a78
14c8379c65a1
156b75ac5078
hex3c8ff38991

260113140113 has 4 divisors (see below), whose sum is σ = 260266239600. Its totient is φ = 259960040628.

The previous prime is 260113140097. The next prime is 260113140163. The reversal of 260113140113 is 311041311062.

It is a happy number.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 260113140113 - 24 = 260113140097 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (260113140163) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 76547195 + ... + 76550592.

It is an arithmetic number, because the mean of its divisors is an integer number (65066559900).

Almost surely, 2260113140113 is an apocalyptic number.

It is an amenable number.

260113140113 is a deficient number, since it is larger than the sum of its proper divisors (153099487).

260113140113 is a wasteful number, since it uses less digits than its factorization.

260113140113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 153099486.

The product of its (nonzero) digits is 432, while the sum is 23.

Adding to 260113140113 its reverse (311041311062), we get a palindrome (571154451175).

The spelling of 260113140113 in words is "two hundred sixty billion, one hundred thirteen million, one hundred forty thousand, one hundred thirteen".

Divisors: 1 1699 153097787 260113140113