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Puzzle -If you are good in maths

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St0neF!sh

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[Caution] Enter Only if you are good in Maths!

[Puzzle]: Breaks apart a 40KG stone into 4 pieces in such a way that using those 4 parts anyone can weight 1KG to 40KG each weights.
Let me describe a little bit more....
Suppose the 4 parts are: A, B, C and D
So,
A+B-C=1KG
B+D-C=2KG
A+D-B=3KG
...............
...............
A+B+C+D=40KG

So, what are the weights of those 4 pieces...??
Note: All pieces are rounded-off in weights, like, each parts are in complete KGs and note in Grams...

Hope you are good enough in maths....huh!

I bet only a mastermind can solve this........

[SOLVED-BY]: Nitish
[ANSWER]: 1,3,9,27
 
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15 comments
[STRIKE]Everytime we gotta use 4 stones only or we can use fewer also?[/STRIKE]

Edit: nevermind, figured it out.
1,3,9,27.
 
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I dont think the question is written out very well...
Unless I just simply don't understand it.

"Breaks apart a 40KG stone into 4 pieces in such a way that using those 4 parts anyone can weight 1KG to 40KG each weights."

Well, that's not right for a start. If one of those 4 pieces weigh 40KG, then where are the other 3? lol
Can't A B C and D all be 10KG's making 40KG ?

I don't know, I just don't get it.
 
A = 6.285 (i.e) A = 44/7
B = 10.571
c = 15.857
D = 7.285 (i.e) D = 51/7

EDIT:
Fully solved in paper :)

So, what are the weights of those 4 pieces...??
Note: All pieces are rounded-off in weights, like, each parts are in complete KGs and note in Grams...
Click to expand...
A = 6.285 ~~ 6
B = 10.571 ~~ 11
c = 15.857 ~~ 16
D = 7.285 ~~ 7

may be i am wrong in rounding off :(



Note:
Even i checked in calculator, value i correct, so round off may wrong

[slide]http://iof5.imgkeep.com/i/00119/gd79boyuqq7k.jpg[/slide]

For checking in calculator:
take 4 x 4 linear equation

(equation 1) => 1A +1B -1C +0D = 1
(equation 2) => 0A +1B -1C +1D = 2
(equation 3) => 1A -1B +0C +1D= 3
(equation 4) => 1A +1B +1C +1D = 40
 
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Ashleyuk1984 said:
I dont think the question is written out very well...
Unless I just simply don't understand it.

"Breaks apart a 40KG stone into 4 pieces in such a way that using those 4 parts anyone can weight 1KG to 40KG each weights."

Well, that's not right for a start. If one of those 4 pieces weigh 40KG, then where are the other 3? lol
Can't A B C and D all be 10KG's making 40KG ?

I don't know, I just don't get it.
Click to expand...

Its common sense that none of them can be 40kg or anything over 37kg also.
You gotta make 4 such weights so that if you add or subtract any weight then you can get any weight from 1-40kg. Try with the answer which i supplied above.

---------- Post added at 10:44 PM ---------- Previous post was at 10:42 PM ----------
torrfriend said:
A = 6.285
B = 10.571
c = 15.857
D = 7.285
Click to expand...

All of them have to be integer weights. Also, if you disregard this rule, you cannot get 2KG out of these weights ( and many more. I just checked for 2).
 
A+B+C+D=40

A+D-B=3
B+D-C=2
A+B-C=1

So,

A+C-2B=1
D=1+A
B=2A-2
C=B+D-2=3A-3

B+C+D=6A-4

7A-4=40
.. 7A=44

Didn't do any proper math for last 10 years :facepalm: But it doesn't go any other way for me :(

If Nitish is correct. Then

A=1
B=6
C=9
D=27

Then, A+B-C=-2 (That is wrong!)
 
Well, this will explain is better.

Google helped me to find out the exact meaning of the question and the solution mentioned here will make the question even clear.

Q. A 40 Kg weight measure broke into 4 pieces in such a way that ,by using one or combinations of them in one or two trays of a balance all weighings could be done from 1 to 40 (no fraction) Please find out the weight of each broken piece.

OR

Q. A boy had a bar of lead that weighed 40 pounds, and he divided it into 4 pieces in such a way as to allow him to weigh any number of pounds from 1 to 40. How did he manage the matter?


SOLUTION



Divide the 40-pound bar into sections of 1, 3, 9, and 27 pounds. Using a two-sided balance scale, place the object to be weighed on the left side. Then place the “added” weights on the right side, and the “subtracted” weights on the left side; if the scale is balanced in the following combinations, then the left-side value is the weight of the object being weighed.
40 = 27 + 9 + 3 + 1 (i. e., if all four weights are on the right side while the object is on the left side of the scale and it is balanced, then the object weighs 40 pounds)
39 = 27 + 9 + 3
38 = 27 + 9 + 3 – 1
37 = 27 + 9 + 1
36 = 27 + 9
35 = 27 + 9 – 1
34 = 27 + 9 – 3 + 1
33 = 27 + 9 – 3
32 = 27 + 9 – 3 – 1
31 = 27 + 3 + 1
30 = 27 + 3
29 = 27 + 3 – 1
28 = 27 + 1
27 = 27
26 = 27 – 1
25 = 27 – 3 + 1
24 = 27 – 3
23 = 27 – 3 – 1
22 = 27 – 9 + 3 + 1
21 = 27 – 9 + 3
20 = 27 – 9 + 3 – 1
19 = 27 – 9 + 1
18 = 27 – 9
17 = 27 – 9 – 1
16 = 27 – 9 – 3 + 1
15 = 27 – 9 – 3
14 = 27 – 9 – 3 – 1
13 = 9 + 3 + 1
12 = 9 + 3
11 = 9 + 3 – 1
10 = 9 + 1
9 = 9
8 = 9 – 1
7 = 9 – 3 + 1
6 = 9 – 3
5 = 9 – 3 – 1
4 = 3 + 1
3 = 3
2 = 3 – 1
1 = 1

How to get the numbers {1 3 9 27}:

The maximum distance between weights must be observed to fill in all the ‘gaps,’ with the smaller weights closer together and the larger weights increasingly far apart. In other words, figure out the smallest units first, then the greatest combination they can reach with all the ‘gaps’ filled in; the difference between the next unknown number minus the sum of the known numbers will be one greater than the sum of the known numbers.

a must = 1, or else the weight 39 could not be achieved.
b – a = 2; therefore b = 3.
since a + b = 4, c – (a + b) = 5, therefore c = 9.
All combinations smaller than a + b + c can now be reached; a + b + c = 13. Therefore d – (a + b + c) must = 14, so d = 27.
Click to expand...

Source - http://www.smart-kit.com/s764/bar-of-lead/
 
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pankaj said:
Well, this will explain is better.

Google helped me to find out the exact meaning of the question and the solution mentioned here will make the question even clear.

Q. A 40 Kg weight measure broke into 4 pieces in such a way that ,by using one or combinations of them in one or two trays of a balance all weighings could be done from 1 to 40 (no fraction) Please find out the weight of each broken piece.

OR

Q. A boy had a bar of lead that weighed 40 pounds, and he divided it into 4 pieces in such a way as to allow him to weigh any number of pounds from 1 to 40. How did he manage the matter?


SOLUTION





Source - http://www.smart-kit.com/s764/bar-of-lead/
Click to expand...
Well...indeed...but then the given conditions are useless :facepalm: Whoever tried those conditions made a wrong conclusion(which are not wrong mathematically though).
 
Yeah, right. I myself failed to understand the question.

I guess Nitish might have searched it via Google because its really very difficult to understand the question in the manner it was explained.
 
LOL!.... If all of you guys didn't understand the question, why don't you ask for more explanation before Nitish replied...

Learn more, before saying "Question is out of Syllabus!"
 
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