6
$\begingroup$

The 12 weights look identical and weigh 1kg,2kg,3kg......12kg.Each mark on the rod is 1m apart. Identify which weight is where and sum of marked weights

enter image description here

$\endgroup$
  • $\begingroup$ Are you sure this is solvable? Two of the weights seem to have to be the same (if we are numbering 1,2,3,... left to right, top to bottom, then the two weights which are equal are 3 and 6 because each is one quarter of the string. Sorry I can't clarify without a diagram) $\endgroup$ – Wen1now Feb 23 '17 at 6:02
  • $\begingroup$ I know it's solvable as I know the answer.I just don't know how it is obtained $\endgroup$ – ULTIMATEGAMER07 Feb 23 '17 at 6:03
  • $\begingroup$ The weights aren't allowed to be the same, are they? $\endgroup$ – Wen1now Feb 23 '17 at 6:04
  • $\begingroup$ Nope 1 to 12 ..Each used once $\endgroup$ – ULTIMATEGAMER07 Feb 23 '17 at 6:05
  • 1
    $\begingroup$ Can you tell us where you got this question? If it's from an ongoing contest or plagiarized, we may have to remove it. $\endgroup$ – Deusovi Feb 23 '17 at 6:22
3
$\begingroup$

Label the weights, from top to bottom then left to right, as A B C D E F G H I J K L.
You have the following relations, as each bar in the system must be level:

3J = 2H+I
3K = L
C = D+2E
F = K+L+2G
2(C+D+E) = 2(F+G+K+L)
A+3(H+I+J) = (C+D+E+F+G+K+L)+3B

Each of these letters must be a unique integer between 1 and 12 inclusive, such that each of those values is used exactly once.

The only solution that satisfies all of these is:

A=10
B=9
C=11
D=7
E=2
F=12
G=4
H=5
I=8
J=6
K=1
L=3

What drastically limits the possibilities is:

2(C+D+E) = 2(F+G+K+L)

As this means ...

C+D+E and F+G+K+L must each be at least 14.
Further confining the values around L=3K and C=D+2E, the possibilities narrow quickly.

At C+D+E=F+G+K+L=20 we find a useful solution.
The other values for H,I,J come quickly as few working choices remain.
At the end, only 9 and 10 are left for A and B, and those are easily placed.

$\endgroup$
  • $\begingroup$ It has never been so painful to get ninja'd... I have been working on this for an hour and just finished it now... Was starting to think about how I could describe the whole mess in my notebook in a post. $\endgroup$ – stack reader Feb 23 '17 at 8:30
  • $\begingroup$ Oh man. I spent the last 15 minutes trying to explain it in a way that was even remotely readable. I know your pain. $\endgroup$ – Rubio Feb 23 '17 at 8:31
  • $\begingroup$ OK so there is no definite way to solve it I guess.I was able to get the answer by trial and error but I posted it here in hopes of finding a way to solve it .Thanks anyway...Nice solution $\endgroup$ – ULTIMATEGAMER07 Feb 23 '17 at 12:48
  • $\begingroup$ How is it that it must be at least 14? Sorry if I'm being stupid $\endgroup$ – ULTIMATEGAMER07 Feb 23 '17 at 12:52
  • $\begingroup$ You have 7 integers, grouped 3 and 4, where the sums of the two groups must be equal. Clearly the minimal values are 1,2,...,6,7 since they must be 1..12 and unique, and thus confined the lowest possible total for all 7 is 28, or 14 per group. (Groups such as, say, 1,2,4,7 + 3,5,6 show that totals of 14 are possible.) You can't get smaller than that without duplicating a value. $\endgroup$ – Rubio Feb 23 '17 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.