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51171433 is a prime number
BaseRepresentation
bin1100001100110…
…1000001101001
310120021210000021
43003031001221
5101044441213
65024440441
71160643565
oct303150151
9116253007
1051171433
1126980945
1215179121
13a7a8671
146b206a5
15475bd8d
hex30cd069

51171433 has 2 divisors, whose sum is σ = 51171434. Its totient is φ = 51171432.

The previous prime is 51171413. The next prime is 51171437. The reversal of 51171433 is 33417115.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 36517849 + 14653584 = 6043^2 + 3828^2 .

It is a cyclic number.

It is not a de Polignac number, because 51171433 - 221 = 49074281 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 51171398 and 51171407.

It is not a weakly prime, because it can be changed into another prime (51171437) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25585716 + 25585717.

It is an arithmetic number, because the mean of its divisors is an integer number (25585717).

Almost surely, 251171433 is an apocalyptic number.

It is an amenable number.

51171433 is a deficient number, since it is larger than the sum of its proper divisors (1).

51171433 is an equidigital number, since it uses as much as digits as its factorization.

51171433 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 1260, while the sum is 25.

The square root of 51171433 is about 7153.4210696701. The cubic root of 51171433 is about 371.2580332846.

Adding to 51171433 its reverse (33417115), we get a palindrome (84588548).

The spelling of 51171433 in words is "fifty-one million, one hundred seventy-one thousand, four hundred thirty-three".