(4!/4)^2 +(4/4)
4x4+4-square root of 4=18
try to find what number times the same numbers equals the number that you have.
101
4 /5
It's really hard to explain in an answer here. See the link for different ways to do it.
The best way to count by fours. If you count by fours you can go to twenty without using other numbers. In the sequence of four, eight, 12, 16, and 20. Sequence counting allows you to follow a pattern to reach the next logical number.
4x4+4-square root of 4=18
I am not aware of a solution using the four basic operations of arithmetic, but otherwise: 129 = [(4^4)/sqrt(4)] + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) * * * * * * * * * * * * No matter how many times you square root it, it will not equal 1 To the OP. I have spend a lot of time on this one. Just wanting to make sure that the correct number is 129 and you are sure you are using only 4 fours. It would work out great if using 5 fours. If you made a mistake, correct it and I'll check back. In the mean time, I'll continue thinking about a solution for 4 fours and 129.
4 times 4 minus 4 minus square root of 4.
44 minus the square root of 4 divided by point 4
try to find what number times the same numbers equals the number that you have.
35=4!+44/4 37=4!+(4!+square root of 4)/square root of 4
4 factorial plus 4 factorial plus the square root of 4 divided by .4
4 + 4 + (4/4).
101
To make 5 using only four 4's, you can either do: 4/4+square root of 4+square root of 4=5 OR: (4x4+4)/4=5 Hope that helped!!
/*Java program to find out square root of a given number * without using any Built-In Functions */ public class SquareRootDemo2 { public static void main(String[] args) { //Number for which square root is to be found double number = -12; //This method finds out the square root findSquareRoot(number); } /*This method finds out the square root without using any built-in functions and displays it */ public static void findSquareRoot(double number) { boolean isPositiveNumber = true; double g1; //if the number given is a 0 if(number==0) { System.out.println("Square root of "+number+" = "+0); } //If the number given is a -ve number else if(number<0) { number=-number; isPositiveNumber = false; } //Proceeding to find out square root of the number double squareRoot = number/2; do { g1=squareRoot; squareRoot = (g1 + (number/g1))/2; } while((g1-squareRoot)!=0); //Displays square root in the case of a positive number if(isPositiveNumber) { System.out.println("Square roots of "+number+" are "); System.out.println("+"+squareRoot); System.out.println("-"+squareRoot); } //Displays square root in the case of a -ve number else { System.out.println("Square roots of -"+number+" are "); System.out.println("+"+squareRoot+" i"); System.out.println("-"+squareRoot+" i"); } } }