Is 2003 a prime number? What are the divisors of 2003?

## Is 2003 a prime number?

Yes, 2003 is a prime number.

Indeed, the definition of a prime numbers is to have only two distinct positive divisors, 1 and itself. A number is a divisor of another number when the remainder of Euclid’s division of the second one by the first one is zero. Concerning the number 2003, the only two divisors are 1 and 2003. Therefore 2003 is a prime number.

As a consequence, 2003 is only a multiple of 1 and 2003.

Therefore year 2003 was a prime year.

Since 2003 is a prime number, 2003 is also a deficient number, that is to say 2003 is a natural integer that is strictly larger than the sum of its proper divisors, i.e., the divisors of 2003 without 2003 itself (that is 1, by definition!).

## Parity of 2003

2003 is an odd number, because it is not evenly divisible by 2.

## Is 2003 a perfect square number?

A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 2003 is about 44.755.

Thus, the square root of 2003 is not an integer, and therefore 2003 is not a square number.

Anyway, 2003 is a prime number, and a prime number cannot be a perfect square.

## What is the square number of 2003?

The square of a number (here 2003) is the result of the product of this number (2003) by itself (i.e., 2003 × 2003); the square of 2003 is sometimes called "raising 2003 to the power 2", or "2003 squared".

The square of 2003 is 4 012 009 because 2003 × 2003 = 20032 = 4 012 009.

As a consequence, 2003 is the square root of 4 012 009.

## Number of digits of 2003

2003 is a number with 4 digits.

## What are the multiples of 2003?

The multiples of 2003 are all integers evenly divisible by 2003, that is all numbers such that the remainder of the division by 2003 is zero. There are infinitely many multiples of 2003. The smallest multiples of 2003 are:

• 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 2003 too, since 0 × 2003 = 0
• 2003: indeed, 2003 is a multiple of itself, since 2003 is evenly divisible by 2003 (we have 2003 / 2003 = 1, so the remainder of this division is indeed zero)
• 4 006: indeed, 4 006 = 2003 × 2
• 6 009: indeed, 6 009 = 2003 × 3
• 8 012: indeed, 8 012 = 2003 × 4
• 10 015: indeed, 10 015 = 2003 × 5
• etc.

## Nearest numbers from 2003

Find out whether some integer is a prime number