Boost the speed of calculating Large Primes with Golang

Following points are used

1. Check whether the last digit of the integer can be divided by 2 or not
2. Check whether the sum of all digits of the integer can be divided by 3 or not
3. Check whether the last digit of the integer can be divided by 5 or not
4. If all the above checks fails Check whether there is an integer from 7 to square root of number which is divide the number

package main import ( "fmt" "math" "strconv" ) func main() { fmt.Printf("275489 is a prime : %t", isPrime(275489)) } func isPrime(num int) bool { if !canBeDivideBy2(num) && !canBeDivideBy3(num) && !canBeDivideBy5(num) { isPrime := true //If the number can not be divided by 2, 3 or 5 divide from 7 to squre root of the number to check the existence of a factor for i := 7; i <= int(math.Sqrt(float64(num))); i++ { if num%i == 0 { isPrime = false break } } return isPrime } return false } // If the last digit can be divided by 2 whole number can be divided by 2 func canBeDivideBy2(num int) bool { numAsStr := strconv.Itoa(num) lastNum, _ := strconv.Atoi(string(numAsStr[len(numAsStr)-1])) return lastNum%2 == 0 } // If the digit sum can be divided by 3 whole number can be divided by 3 func canBeDivideBy3(num int) bool { numAsStr := strconv.Itoa(num) sum := 0 for i := 0; i < len(numAsStr); i++ { iThNum, _ := strconv.Atoi(string(numAsStr[i])) sum += iThNum } return sum%3 == 0 } //If the last digit can be divided by 5 whole number can be divided by 5 func canBeDivideBy5(num int) bool { numAsStr := strconv.Itoa(num) lastNum, _ := strconv.Atoi(string(numAsStr[len(numAsStr)-1])) return lastNum%5 == 0 }