Math Problem 5! Help Appreciated

[Dunzo]

Hey guys! I have a problem of the week for my math class I haven't really looked at it but if you all could help it would be greatly appreciated! If you agree with one person's answer please post saying so. Thanks!

WEIGHING WEIGHTS

Every integer weight from 1 lb. to 40 lbs. can be weighed using a balance scale with only four different weights. Determine these four weights. Explain how you would weigh 19 lbs.

[HavokD]

The different weights are 1, 3, 9 and 27. To weigh 19 lbs, immagine you have an empty scale. On the left side you put your 19 lbs weight, on the right you must put the 27 lbs weight. We need to balance our scale, so we add 9 lbs on the left side of the scale. That gives us 19+9=28 lbs on the left side, and 27 lbs on the right. So finally we add the 1 lb weight to the right side. Now the scale is balanced, with 28 lbs on each side, but knowing that we are weighing only 19 lbs.

As for the numbers, I had already known the answer to this problem since it's very known. I didnt solve it, we were just given the numbers long time ago in physics class if I recall correctly. I can try and ellaborate a decent answer to it if you would like me to, tho.

G luck xD

Edit: Actually I just put myself to work and I got to the answer, tho I cant say its a proper solution, its a good procedure:

First of all I knew that w + x + y + z =40, which would have to be the maximum number I could have.

Now, I also knew a 1 lb weight would be needed, by logic, seeing that my lowest number was going to be 1.

Then I started adding up numbers to try to get a 2, 3, etc, till 40 (if I had to) and see if there was a pattern at all. And I found it quite fast xD

So for a 2:

2= 1+1 -> Discarded, waste of weight to have two of 1 lb
2= 3-1
2= 4-2 -> Discarded, I have another 2 in here.
2= 5-3
2 =6-4

Now for a 3:

3= 4-1
3= 5-2 -> We are supposing we dont have a two in our weights, so this is discarded
3= 6-3-> Discarded, we cant have another three.
3= 7-4

For a 4:

4= 3+1
4= 5-1
4= 6-2 -> If we keep our supposition, we cant have a two.
4= 7-3
4= 8-4 -> Discarded, you should know why right now xD
4= 9-5

Now after this, I tried to find some patterns, some numbers that repeated themselves, and tried to use them more for the next numbers.
This numbers were 1 (by default) and 3. These numbers were used combined to get 1 thru 4. So next I tried to use these numbers to form a 5, but ommitting 2, 4, 5 and 6 since they werent useful before:

5= 3+1+1 -> Discarded, we cant have two same weights by logical reasons.
5= 3-1+3-> Dscarded by the same reason as before
5= 7-3+1-> This one sounds ok, remember I already burned 2, 4, 5 and 6.
5= 9-3-1-> This one sounds alright aswell.
5= 8-3 -> Same with this
5= 10-3-2 -> Already burned the 2

I stopped here. I had some good leads, now I tried to reproduce the next numbers using 1, 3 (which I could say by now were 2 of our weights) and a 7, 9 or an 8.

6= 7-1 --> Good
6= 8-2 -> We already dscarded number 2, this one gets discarded aswell
6= 9-3 -> Good
6= 8-3+1 -> Good

Well, either 7, 8 or 9 are fine with this. We continue:

7= 7 -> Dumb, I know, but still if we suppose our third weight is a 7 this should work.
7= 8-1 -> Good
7= 9-2 -> We discarded 2, this wont do
7= 9-3-1 -> Good

We keep 7, 8 and 9. Please be aware that we only want to reproduce operations using 1, 3, and either a 7, 8 or 9. We can use up to three numbers, or just one. We do this to ensure we are on the right path.

Next one

8= 9-1
8= 8
8= 7+1

We continue:

9= 9
9= 8+1
9= 7+2 -> Discarded, you know why

Theres no other operation that involves a 7 and doesnt involve any number we already burned. So good-bye to Mr 7
We continue:

10= 9+1 -> Good
10= 8+2 -> Bad, we already burned our 2.
10= 8+3-1 -> Good

We go on

11= 9+2 -> Cant do this
11= 9+3-1 -> Good
11= 8+3 -> Good

Go on:

12= 9+3 -> Good
12= 8+3+1 -> Good

Go on

13= 9+3+1 -> Good

Sadly we cant make a 13 using an 8 and our other available numbers. So we could either assume there is another number right now, or that 9, 3 and 1 are correct numbers for our weights.

I will procede with option number 2.

To make a 14, now, we have a problem, the maximum number we could have gotten was 9 + 3 +1=13, so now we add a variable:
We try to make the combintions needed, but since we know our next number must be higher than 9, the only viable options are:

14= X-9
14= X-3
14= X-1
14= X-9-3
14= X-9-1
14= X-3-1
14 = X - 9 - 3 - 1

This yields 7 different values for our next and last Weight (if our supposition is correct).
X= 23, 17, 15, 26, 24, 18, 27

Now I will try to recreate the next number using 1, 3, 9 and X.

15= 23-9+1 -> Good
15= 17-3+1 -> Good
15= 15 -> Obviously Good
15= 26-9-3+1 -> Good
15= 24-9 -> Good
15= 18-3 -> Good
15= 27 - 9-3 ->Good

We keep our numbers, we go on to 16

16= 23-9+3-1 -> Good
16= 17-1
16= 15+1
16= 26-9-1
16= 24-9+1
16= 18-3+1
16= 27-9-3+1

This is getting boring, and I suspect we will be able to reproduce the next few numbers using this available numbers. I will jump to 35 (randomly, you could try any other number and you will find the same answer) and see how well it fares:

35= 23+9+3
35= 17 + 9 + 3 + 1 --> DIscarded, maximum value that 17 can get is 30,and so it gets burned
35= 15+9+3+1 --> Same as with 17
35= 26+9
35= 24+9+3-1
35=18+9+3+1 -> Discarded, same as with 15 and 17.
35= 27+9-1

We are keeping now 1, 3, 9 and X=23, 26, 24 and 27.

Lets jump to 38:
38= 23+9+3+1 -> Discarded, it is 36
38= 26+9+3
35= 24+9+3+1 -> Discarded, plus 24 sucks
35= 27+9+3-1

We now have X=26, 27

Lets jump to 40!
40= 26+9+3+1 -> DISCARDED, this yields 39, OH HOW CLOSE!
40= 27+9+3+1 -> Win

So we have come to the result that the weights must be 1, 3, 9 and 27.

Edit #2: Actually after getting the different values our X could yield you could have technically jumped to number 40 and try to get the right X value without having to do the rest of the numbers, which kills alot of time.

[Dunzo]

if u could it would be lovely :P

[HavokD]

Well just finished xD

[samzo]

dunzo's teacher must think he's a genius

[Olzone]

unusual to surf into blackdtools and see it's a math forum ;o

[Crille]

Aren't you supposed to solve these by yourself?

[Olzone]

Aren't you supposed to solve these by yourself?

He should, but HavokD is kinda a "Teacher" for him to give him examples and figure it out for himself.. try use the Private Message function and if he can't help you.. all of us would be to your help :P?

[iBotzer]

Dam HavokD you would help me on my Geom homework x_x

Like Seriously ! Haha

[HavokD]

If you want me to xD but Im not much into geometry, tho I have a couple of books here at my house that I can use as reference if needed x.X I once attended some "conferences" about geometrical problems, and I must say thats one of my many weaknesses X.X

[iBotzer]

Sweet !

I really need help on Proofs and Statements xD

[Tony32]

The only thing here I think is annoying, is that you create a new thread for every math problem. Why? <.<

[X117]

to keep this forum aliveeee! aliveee!