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211132013 is a prime number
BaseRepresentation
bin11001001010110…
…01111001101101
3112201021121200012
430211121321231
5413022211023
632541143005
75142410141
oct1445317155
9481247605
10211132013
11a91a6674
125a85aa65
13349841b1
142007d221
1513807978
hexc959e6d

211132013 has 2 divisors, whose sum is σ = 211132014. Its totient is φ = 211132012.

The previous prime is 211131983. The next prime is 211132057. The reversal of 211132013 is 310231112.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 194379364 + 16752649 = 13942^2 + 4093^2 .

It is a cyclic number.

It is not a de Polignac number, because 211132013 - 216 = 211066477 is a prime.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (211132063) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 105566006 + 105566007.

It is an arithmetic number, because the mean of its divisors is an integer number (105566007).

Almost surely, 2211132013 is an apocalyptic number.

It is an amenable number.

211132013 is a deficient number, since it is larger than the sum of its proper divisors (1).

211132013 is an equidigital number, since it uses as much as digits as its factorization.

211132013 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 36, while the sum is 14.

The square root of 211132013 is about 14530.3824106594. The cubic root of 211132013 is about 595.4583131726.

Adding to 211132013 its reverse (310231112), we get a palindrome (521363125).

Subtracting 211132013 from its reverse (310231112), we obtain a palindrome (99099099).

The spelling of 211132013 in words is "two hundred eleven million, one hundred thirty-two thousand, thirteen".