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110110001153 is a prime number
BaseRepresentation
bin110011010001100010…
…0010100100000000001
3101112012202101121120212
41212203010110200001
53301001130014103
6122330035243505
710645425365345
oct1464304244001
9345182347525
10110110001153
1142774252659
121940b749595
13a4ca232949
145487a59625
152ce6ac57d8
hex19a3114801

110110001153 has 2 divisors, whose sum is σ = 110110001154. Its totient is φ = 110110001152.

The previous prime is 110110001101. The next prime is 110110001191. The reversal of 110110001153 is 351100011011.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 66741105649 + 43368895504 = 258343^2 + 208252^2 .

It is a cyclic number.

It is not a de Polignac number, because 110110001153 - 224 = 110093223937 is a prime.

It is a super-2 number, since 2×1101100011532 (a number of 23 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (110110009153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55055000576 + 55055000577.

It is an arithmetic number, because the mean of its divisors is an integer number (55055000577).

Almost surely, 2110110001153 is an apocalyptic number.

It is an amenable number.

110110001153 is a deficient number, since it is larger than the sum of its proper divisors (1).

110110001153 is an equidigital number, since it uses as much as digits as its factorization.

110110001153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15, while the sum is 14.

Adding to 110110001153 its reverse (351100011011), we get a palindrome (461210012164).

The spelling of 110110001153 in words is "one hundred ten billion, one hundred ten million, one thousand, one hundred fifty-three".