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5325113200049 is a prime number
BaseRepresentation
bin100110101111101100101…
…1110110000110110110001
3200212002001021221220020122
41031133121132300312301
51144221312434400144
615154153452420025
71056504161236256
oct115373136606661
920762037856218
105325113200049
111773405637745
12720061777615
132c8204a12156
14145a45cdb42d
15937b99b37ee
hex4d7d97b0db1

5325113200049 has 2 divisors, whose sum is σ = 5325113200050. Its totient is φ = 5325113200048.

The previous prime is 5325113200021. The next prime is 5325113200061. The reversal of 5325113200049 is 9400023115235.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 5059656399424 + 265456800625 = 2249368^2 + 515225^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-5325113200049 is a prime.

It is a super-2 number, since 2×53251132000492 (a number of 26 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 5325113199991 and 5325113200018.

It is not a weakly prime, because it can be changed into another prime (5325110200049) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2662556600024 + 2662556600025.

It is an arithmetic number, because the mean of its divisors is an integer number (2662556600025).

Almost surely, 25325113200049 is an apocalyptic number.

It is an amenable number.

5325113200049 is a deficient number, since it is larger than the sum of its proper divisors (1).

5325113200049 is an equidigital number, since it uses as much as digits as its factorization.

5325113200049 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 32400, while the sum is 35.

The spelling of 5325113200049 in words is "five trillion, three hundred twenty-five billion, one hundred thirteen million, two hundred thousand, forty-nine".