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63104551713347 is a prime number
BaseRepresentation
bin11100101100100101011000…
…11000110111101001000011
322021102201212201011010112212
432112102230120313221003
531232401110214311342
6342113455244012335
716202103422203364
oct1626225430675103
9267381781133485
1063104551713347
111911a543345551
1270b21028310ab
13292996cbcc821
1411823c2dbbb6b
15746762e37182
hex3964ac637a43

63104551713347 has 2 divisors, whose sum is σ = 63104551713348. Its totient is φ = 63104551713346.

The previous prime is 63104551713331. The next prime is 63104551713353. The reversal of 63104551713347 is 74331715540136.

63104551713347 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 63104551713347 - 24 = 63104551713331 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 63104551713295 and 63104551713304.

It is not a weakly prime, because it can be changed into another prime (63104551713947) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 31552275856673 + 31552275856674.

It is an arithmetic number, because the mean of its divisors is an integer number (31552275856674).

Almost surely, 263104551713347 is an apocalyptic number.

63104551713347 is a deficient number, since it is larger than the sum of its proper divisors (1).

63104551713347 is an equidigital number, since it uses as much as digits as its factorization.

63104551713347 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3175200, while the sum is 50.

The spelling of 63104551713347 in words is "sixty-three trillion, one hundred four billion, five hundred fifty-one million, seven hundred thirteen thousand, three hundred forty-seven".