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BaseRepresentation
bin10111101010001
3121121122
42331101
5341423
6132025
750213
oct27521
917548
1012113
119112
127015
13568a
1445b3
1538c8
hex2f51

12113 has 2 divisors, whose sum is σ = 12114. Its totient is φ = 12112.

The previous prime is 12109. The next prime is 12119. The reversal of 12113 is 31121.

It is a m-pointer prime, because the next prime (12119) can be obtained adding 12113 to its product of digits (6).

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 9409 + 2704 = 97^2 + 52^2 .

12113 is a truncatable prime.

It is an emirp because it is prime and its reverse (31121) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 12113 - 22 = 12109 is a prime.

It is a Chen prime.

It is a plaindrome in base 13.

It is a self number, because there is not a number n which added to its sum of digits gives 12113.

It is not a weakly prime, because it can be changed into another prime (12119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6056 + 6057.

It is an arithmetic number, because the mean of its divisors is an integer number (6057).

212113 is an apocalyptic number.

It is an amenable number.

12113 is a deficient number, since it is larger than the sum of its proper divisors (1).

12113 is an equidigital number, since it uses as much as digits as its factorization.

12113 is an evil number, because the sum of its binary digits is even.

The product of its digits is 6, while the sum is 8.

The square root of 12113 is about 110.0590750461. The cubic root of 12113 is about 22.9659230713.

Adding to 12113 its reverse (31121), we get a palindrome (43234).

It can be divided in two parts, 12 and 113, that added together give a cube (125 = 53).

The spelling of 12113 in words is "twelve thousand, one hundred thirteen".