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138063583251947 is a prime number
BaseRepresentation
bin11111011001000101101110…
…110101100000110111101011
3200002211202020111111112222102
4133121011232311200313223
5121044013204303030242
61205345311243444015
741036523204354263
oct3731055665406753
9602752214445872
10138063583251947
113aa9a489487771
121359979282660b
135c064560a1442
1426144551607a3
1510e653d615e32
hex7d916ed60deb

138063583251947 has 2 divisors, whose sum is σ = 138063583251948. Its totient is φ = 138063583251946.

The previous prime is 138063583251943. The next prime is 138063583251949. The reversal of 138063583251947 is 749152385360831.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 138063583251947 - 22 = 138063583251943 is a prime.

Together with 138063583251949, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (138063583251943) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 69031791625973 + 69031791625974.

It is an arithmetic number, because the mean of its divisors is an integer number (69031791625974).

It is a 2-persistent number, because it is pandigital, and so is 2⋅138063583251947 = 276127166503894, but 3⋅138063583251947 = 414190749755841 is not.

Almost surely, 2138063583251947 is an apocalyptic number.

138063583251947 is a deficient number, since it is larger than the sum of its proper divisors (1).

138063583251947 is an equidigital number, since it uses as much as digits as its factorization.

138063583251947 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 130636800, while the sum is 65.

The spelling of 138063583251947 in words is "one hundred thirty-eight trillion, sixty-three billion, five hundred eighty-three million, two hundred fifty-one thousand, nine hundred forty-seven".