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73544050253389 is a prime number
BaseRepresentation
bin10000101110001101001110…
…111111011001111001001101
3100122101202001022102111010111
4100232031032333121321031
534114421203331102024
6420225400440040021
721330245060525302
oct2056151677317115
9318352038374114
1073544050253389
1121484947095691
1282b93b5763011
1332062358a2161
1414237a00250a9
158780b2204a94
hex42e34efd9e4d

73544050253389 has 2 divisors, whose sum is σ = 73544050253390. Its totient is φ = 73544050253388.

The previous prime is 73544050253273. The next prime is 73544050253413. The reversal of 73544050253389 is 98335205044537.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 72520041856900 + 1024008396489 = 8515870^2 + 1011933^2 .

It is a cyclic number.

It is not a de Polignac number, because 73544050253389 - 229 = 73543513382477 is a prime.

It is a super-3 number, since 3×735440502533893 (a number of 43 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (73544050253489) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 36772025126694 + 36772025126695.

It is an arithmetic number, because the mean of its divisors is an integer number (36772025126695).

Almost surely, 273544050253389 is an apocalyptic number.

It is an amenable number.

73544050253389 is a deficient number, since it is larger than the sum of its proper divisors (1).

73544050253389 is an equidigital number, since it uses as much as digits as its factorization.

73544050253389 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 54432000, while the sum is 58.

The spelling of 73544050253389 in words is "seventy-three trillion, five hundred forty-four billion, fifty million, two hundred fifty-three thousand, three hundred eighty-nine".