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51002205433 is a prime number
BaseRepresentation
bin101111011111111101…
…111110010011111001
311212122102122111000121
4233133331332103321
51313423031033213
635232522133241
73453611410324
oct573775762371
9155572574017
1051002205433
111a6a2462903
129a74651221
134a6a59c9aa
14267b8937bb
1514d785e88d
hexbdff7e4f9

51002205433 has 2 divisors, whose sum is σ = 51002205434. Its totient is φ = 51002205432.

The previous prime is 51002205409. The next prime is 51002205473. The reversal of 51002205433 is 33450220015.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 50857917289 + 144288144 = 225517^2 + 12012^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-51002205433 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 51002205398 and 51002205407.

It is not a weakly prime, because it can be changed into another prime (51002205403) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25501102716 + 25501102717.

It is an arithmetic number, because the mean of its divisors is an integer number (25501102717).

Almost surely, 251002205433 is an apocalyptic number.

It is an amenable number.

51002205433 is a deficient number, since it is larger than the sum of its proper divisors (1).

51002205433 is an equidigital number, since it uses as much as digits as its factorization.

51002205433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3600, while the sum is 25.

Adding to 51002205433 its reverse (33450220015), we get a palindrome (84452425448).

The spelling of 51002205433 in words is "fifty-one billion, two million, two hundred five thousand, four hundred thirty-three".