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BaseRepresentation
bin1011100001111
322002122
41130033
5142103
643155
723132
oct13417
98078
105903
114487
1234bb
1328c1
142219
151b38
hex170f

5903 has 2 divisors, whose sum is σ = 5904. Its totient is φ = 5902.

The previous prime is 5897. The next prime is 5923. The reversal of 5903 is 3095.

5903 is nontrivially palindromic in base 7.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 5903 - 26 = 5839 is a prime.

It is a Sophie Germain prime.

It is a Chen prime.

It is a plaindrome in base 12.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (5923) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2951 + 2952.

It is an arithmetic number, because the mean of its divisors is an integer number (2952).

5903 is a deficient number, since it is larger than the sum of its proper divisors (1).

5903 is an equidigital number, since it uses as much as digits as its factorization.

5903 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 135, while the sum is 17.

The square root of 5903 is about 76.8309833335. The cubic root of 5903 is about 18.0727508334.

Subtracting from 5903 its sum of digits (17), we obtain a triangular number (5886 = T108).

Adding to 5903 its reverse (3095), we get a palindrome (8998).

It can be divided in two parts, 590 and 3, that multiplied together give a triangular number (1770 = T59).

The spelling of 5903 in words is "five thousand, nine hundred three".