Math quiz [solved]

Mustafa Ismail Mustafa

OK, let's see if you guys can work this one out :) The numbers: 1, 2, 3, 4, 5 and the operations +, -, *, / can be used in any order (parenthesis are free to control the order of evaluation) All the preceding mentioned numbers and operations can be used in any order once and only once to provide the result of 28. Remember, each of the operations can be used once only and each of the numbers can be used once only. You have to use them all to achieve your goal. You may begin :) [solution] Credited to aspdotnetdev (4 - (1/2)) * (5 + 3) = 28. Points awarded. Would you like me to make this a feature? Sorta like the CCC but for us more Mathematically as opposed to being "constantly high on peyote in order to be able to answer the CCC"?

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modified on Friday, October 15, 2010 2:52 PM

Chris Meech · modified on Friday, October 15, 2010 2:52 PM

(2+3)*5+4-1 = 28. My kids and I used to play this game all the time. :) [EDIT] Just saw your rule about using all operators and only once. Not familiar with that one. :doh: :doh: :doh:

Chris Meech I am Canadian. [heard in a local bar] In theory there is no difference between theory and practice. In practice there is. [Yogi Berra]

Rajesh R Subramanian · modified on Friday, October 15, 2010 2:52 PM

Math on a Friday night? Get stuffed man. I'm here after a heavy meal, and some ripe red bananas are on the table. After I savour on them, I'll to get some good rest. G'Nite. :)

There are some really weird people on this planet - MIM.

SinghUlarity

You used + twice ;P

I are n00b.

SinghUlarity · modified on Friday, October 15, 2010 2:52 PM

(5-1) * (4 + (3 / (11%2))) you never said i couldn't introduce new numbers and operators ;P

I are n00b.

AspDotNetDev

SinghUlarity... wrote:

introduce new numbers

28 = 28 ;)

[Forum Guidelines]

leppie · modified on Friday, October 15, 2010 2:52 PM

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

Took 4 seconds. How I got there: Started with biggest number first.

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

SinghUlarity

To be wrong :-D

I are n00b.

leppie

SinghUlarity... wrote:

To be wrong

FFS, I missed reading one word! :doh: Damnit, now it is a lot harder!

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

AspDotNetDev · modified on Friday, October 15, 2010 2:52 PM

4 * (5 + 2)

[Forum Guidelines]

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 :)

[Forum Guidelines]

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." :((

[Forum Guidelines]

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

[Forum Guidelines]

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 · modified on Friday, October 15, 2010 2:52 PM

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.

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