It is possible to find out using mathematical methods whether a given integer is a prime number or not.
For 953, the answer is: yes, 953 is a prime number because it has only two distinct divisors: 1 and itself (953).
As a consequence, 953 is only a multiple of 1 and 953..
Since 953 is a prime number, 953 is also a deficient number, that is to say 953 is a natural integer that is strictly larger than the sum of its proper divisors, i.e., the divisors of 953 without 953 itself (that is 1, by definition!).
Parity of 953
953 is an odd number, because it is not evenly divisible by 2.
Is 953 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 953 is about 30.871.
Thus, the square root of 953 is not an integer, and therefore 953 is not a square number.
Anyway, 953 is a prime number, and a prime number cannot be a perfect square.
What is the square number of 953?
The square of a number (here 953) is the result of the product of this number (953) by itself (i.e., 953 × 953); the square of 953 is sometimes called "raising 953 to the power 2", or "953 squared".
As a consequence, 953 is the square root of 908 209.
Number of digits of 953
953 is a number with 3 digits.
What are the multiples of 953?
The multiples of 953 are all integers evenly divisible by 953, that is all numbers such that the remainder of the division by 953 is zero. There are infinitely many multiples of 953. The smallest multiples of 953 are:
- 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 953 too, since 0 × 953 = 0
- 953: indeed, 953 is a multiple of itself, since 953 is evenly divisible by 953 (we have 953 / 953 = 1, so the remainder of this division is indeed zero)
- 1 906: indeed, 1 906 = 953 × 2
- 2 859: indeed, 2 859 = 953 × 3
- 3 812: indeed, 3 812 = 953 × 4
- 4 765: indeed, 4 765 = 953 × 5
How to determine whether an integer is a prime number?
To determine the primality of a number, several algorithms can be used. The most naive technique is to test all divisors strictly smaller to the number of which we want to determine the primality (here 953). First, we can eliminate all even numbers greater than 2 (and hence 4, 6, 8…). Then, we can stop this check when we reach the square root of the number of which we want to determine the primality (here the square root is about 30.871). Historically, the sieve of Eratosthenes (dating from the Greek mathematics) implements this technique in a relatively efficient manner.
More modern techniques include the sieve of Atkin, probabilistic algorithms, and the cyclotomic AKS test.