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12012110000153 is a prime number
BaseRepresentation
bin1010111011001100100110…
…0011001011000000011001
31120112100100121011211222122
42232303021203023000121
53033301300130001103
641314142005231025
72346563166221264
oct256631143130031
946470317154878
1012012110000153
113911341a3819a
12142003b402475
13691977589853
142d7564071cdb
1515c6e116db38
hexaecc98cb019

12012110000153 has 2 divisors, whose sum is σ = 12012110000154. Its totient is φ = 12012110000152.

The previous prime is 12012110000041. The next prime is 12012110000159. The reversal of 12012110000153 is 35100001121021.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11129643276544 + 882466723609 = 3336112^2 + 939397^2 .

It is a cyclic number.

It is not a de Polignac number, because 12012110000153 - 224 = 12012093222937 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is not a weakly prime, because it can be changed into another prime (12012110000159) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6006055000076 + 6006055000077.

It is an arithmetic number, because the mean of its divisors is an integer number (6006055000077).

Almost surely, 212012110000153 is an apocalyptic number.

It is an amenable number.

12012110000153 is a deficient number, since it is larger than the sum of its proper divisors (1).

12012110000153 is an equidigital number, since it uses as much as digits as its factorization.

12012110000153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 60, while the sum is 17.

Adding to 12012110000153 its reverse (35100001121021), we get a palindrome (47112111121174).

The spelling of 12012110000153 in words is "twelve trillion, twelve billion, one hundred ten million, one hundred fifty-three", and thus it is an aban number.