Is 8423 a prime number? What are the divisors of 8423?

## Is 8423 a prime number?

Yes, 8423 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 8423, the only two divisors are 1 and 8423. Therefore 8423 is a prime number.

As a consequence, 8423 is only a multiple of 1 and 8423.

Since 8423 is a prime number, 8423 is also a deficient number, that is to say 8423 is a natural integer that is strictly larger than the sum of its proper divisors, i.e., the divisors of 8423 without 8423 itself (that is 1, by definition!).

## Parity of 8423

8423 is an odd number, because it is not evenly divisible by 2.

## Is 8423 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 8423 is about 91.777.

Thus, the square root of 8423 is not an integer, and therefore 8423 is not a square number.

Anyway, 8423 is a prime number, and a prime number cannot be a perfect square.

## What is the square number of 8423?

The square of a number (here 8423) is the result of the product of this number (8423) by itself (i.e., 8423 × 8423); the square of 8423 is sometimes called "raising 8423 to the power 2", or "8423 squared".

The square of 8423 is 70 946 929 because 8423 × 8423 = 84232 = 70 946 929.

As a consequence, 8423 is the square root of 70 946 929.

## Number of digits of 8423

8423 is a number with 4 digits.

## What are the multiples of 8423?

The multiples of 8423 are all integers evenly divisible by 8423, that is all numbers such that the remainder of the division by 8423 is zero. There are infinitely many multiples of 8423. The smallest multiples of 8423 are:

• 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 8423 too, since 0 × 8423 = 0
• 8423: indeed, 8423 is a multiple of itself, since 8423 is evenly divisible by 8423 (we have 8423 / 8423 = 1, so the remainder of this division is indeed zero)
• 16 846: indeed, 16 846 = 8423 × 2
• 25 269: indeed, 25 269 = 8423 × 3
• 33 692: indeed, 33 692 = 8423 × 4
• 42 115: indeed, 42 115 = 8423 × 5
• etc.

## Nearest numbers from 8423

Find out whether some integer is a prime number