Search a number
-
+
12956625137 is a prime number
BaseRepresentation
bin11000001000100011…
…00110100011110001
31020102222011202100002
430010101212203301
5203013344001022
65541401235345
7636035332211
oct140421464361
936388152302
1012956625137
11554974a265
122617192b55
1312b63c5cc1
148acab8241
1550c735092
hex3044668f1

12956625137 has 2 divisors, whose sum is σ = 12956625138. Its totient is φ = 12956625136.

The previous prime is 12956625107. The next prime is 12956625139. The reversal of 12956625137 is 73152665921.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11164880896 + 1791744241 = 105664^2 + 42329^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-12956625137 is a prime.

Together with 12956625139, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 12956625091 and 12956625100.

It is not a weakly prime, because it can be changed into another prime (12956625139) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6478312568 + 6478312569.

It is an arithmetic number, because the mean of its divisors is an integer number (6478312569).

Almost surely, 212956625137 is an apocalyptic number.

It is an amenable number.

12956625137 is a deficient number, since it is larger than the sum of its proper divisors (1).

12956625137 is an equidigital number, since it uses as much as digits as its factorization.

12956625137 is an evil number, because the sum of its binary digits is even.

The product of its digits is 680400, while the sum is 47.

The spelling of 12956625137 in words is "twelve billion, nine hundred fifty-six million, six hundred twenty-five thousand, one hundred thirty-seven".