Math quiz [solved]

[AspDotNetDev]

4 * (5 + 2)

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[leppie]

Very sneaky! Finding the loophole. The OP should have used must instead of can :)

xacc.ide
IronScheme - 1.0 RC 1 - out now!
((λ (x) `(,x ',x)) '(λ (x) `(,x ',x))) The Scheme Programming Language – Fourth Edition

[AspDotNetDev]

Hint: 4 * 7 = 28 :)

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[SinghUlarity]

And that's how the clients end up with a Monkey when they order a Giraffe :laugh:

I are n00b.

[AspDotNetDev]

Just saw this bit: "You have to use them all to achieve your goal." :((

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[SinghUlarity]

You have to use them all to achieve your goal. :rolleyes:

I are n00b.

[Henry Minute]

Mustafa did say "You have to use them all". Last sentence.

Henry Minute Do not read medical books! You could die of a misprint. - Mark Twain Girl: (staring) "Why do you need an icy cucumber?" “I want to report a fraud. The government is lying to us all.”

[AspDotNetDev]

Perhaps the OP was mistaken. ;P

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[Mustafa Ismail Mustafa]

Nope, not mistaken :)

If the post was helpful, please vote, eh! Current activities: Playing Star Craft II. Don't bother me, eh? Now and forever, defiant to the end. What is Multiple Sclerosis[^]?

[AspDotNetDev]

I do not think a solution exists. The closest my computer could get was 27. :((

using System.Collections.Generic;
using System.Windows.Forms;

namespace WindowsFormsApplication1
{
public partial class Form1 : Form
{

    delegate double Op(double num1, double num2);

    public Form1()
    {
        InitializeComponent();

        List<Op> ops = new List<Op> { new Op(M), new Op(D), new Op(A), new Op(S) };

        List<int> opOrder = new List<int> { 0, 0, 0, 0 };

        List<double> results = new List<double>();
        for (int op1 = 0; op1 < 4; op1++)
        {
            for (int op2 = 0; op2 < 4; op2++)
            {
                for (int op3 = 0; op3 < 4; op3++)
                {
                    for (int op4 = 0; op4 < 4; op4++)
                    {
                        for (int n1 = 1; n1 <= 5; n1++)
                        {
                            for (int n2 = 1; n2 <= 5; n2++)
                            {
                                for (int n3 = 1; n3 <= 5; n3++)
                                {
                                    for (int n4 = 1; n4 <= 5; n4++)
                                    {
                                        for (int n5 = 1; n5 <= 5; n5++)
                                        {
                                            bool valid = true;
                                            List<int> opTest = new List<int> { op1, op2, op3, op4 };
                                            List<int> numTest = new List<int> { n1, n2, n3, n4, n5 };
                                            for (int i = 0; i < 4; i++)
                                            {
                                                for (int j = 0; j < 4; j++)
                                                {
                                                    if (i != j)
                                                    {
                                                        if (opTest\[i\] == opTest\[j\])
                                                        {
                                                            valid = false;
                                                        }
                                                    }

[Ian Shlasko]

static void Main(string[] args)
{
int[] nums = new int[] { 1,1,1,1,1};

while (true)
{
    Signs\[\] signs = new Signs\[4\];

    while (true)
    {
        if (nums.Sum() == 15)
        {

            double x = nums\[0\];
            for (int idx = 1; idx < 5; idx++)
                switch (signs\[idx - 1\])
                {
                    case Signs.Plus: x += nums\[idx\]; break;
                    case Signs.Minus: x -= nums\[idx\]; break;
                    case Signs.Div: x /= Convert.ToDouble(nums\[idx\]); break;
                    case Signs.Times: x \*= nums\[idx\]; break;
                }

            if (x == 28 && nums.Distinct().Count() != 5 && signs.Distinct().Count() == 4)
            {
                // Breakpoint here... What?  I'm lazy.
            }

        }

        signs\[3\]++;
        for (int idx = 3; idx > 0; idx--)
            if ((int)signs\[idx\] == 4)
            {
                signs\[idx\] = Signs.Plus;
                signs\[idx - 1\]++;
            }
        if ((int)signs\[0\] > 3) break;

        Console.Write('.');
    }



    nums\[4\]++;
    for (int idx = 4; idx > 0; idx--)
    {
        if (nums\[idx\] > 5)
        {
            nums\[idx\] = 1;
            nums\[idx - 1\]++;
        }
    }
    if (nums\[0\] > 5) break;
}

}

private enum Signs : int
{
Plus = 0,
Minus = 1,
Times = 2,
Div = 3
}

But this isn't inclusive... It wouldn't try something like (5+1)*(4-3)/2, because you can't do that sequentially.

Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels)

[AspDotNetDev]

Dang, I think you're right. I didn't account for parens. Will attempt to modify...

[Forum Guidelines]

[SinghUlarity]

While we kill ourselves over this problem, Mustafa is out somewhere enjoying his weekend. This is all a set-up :suss:

I are n00b.

[GrumbleDuke]

Using integer math: 4*(5+2)-1/3 = 28

[Ian Shlasko]

But he said numbers, not integers, so you can't just round it :)

Proud to have finally moved to the A-Ark. Which one are you in?
Author of the Guardians Saga (Sci-Fi/Fantasy novels)

[leppie]

or I downvote you! ;P This is not possible I think. I have run an exhaustive search over my brain, and now I am tired.

xacc.ide
IronScheme - 1.0 RC 1 - out now!
((λ (x) `(,x ',x)) '(λ (x) `(,x ',x))) The Scheme Programming Language – Fourth Edition

[AspDotNetDev]

If there is a solution, I must find it. If there is not, I must prove that!

[Forum Guidelines]

[leppie]

Practice on Fermat's Last Theorem then ;P

xacc.ide
IronScheme - 1.0 RC 1 - out now!
((λ (x) `(,x ',x)) '(λ (x) `(,x ',x))) The Scheme Programming Language – Fourth Edition

[AspDotNetDev]

(4 - 1 / 2) * (5 + 3)

[Forum Guidelines]

[Sandeep Mewara]

Mustafa Ismail Mustafa wrote:

parenthesis are free to control the order of evaluation

((2(3+5))*(2/1)) - 4 :rolleyes: UPDATE: [(2){3*5+1}]-4 :mad: on self... UPDATE 2: This looks good.... (5+3)*(4-(1/2)) derived from 28=56/2=8*7/2.....

modified on Friday, October 15, 2010 2:45 PM