Doubling up
Five schoolchildren were weighed in pairs. Their (combined) weights were:
129, 125, 124, 123, 122,121, 120, 118, 116 and 114 lbs. What were their individual weights?
129, 125, 124, 123, 122,121, 120, 118, 116 and 114 lbs. What were their individual weights?
Labels: mathschallenge
posted by Chris at
8:28 PM
32 Comments:
Hi Chris ... don't understand this.
It appears you already gave individual weights ...were there 5 or 5 pairs or ... need calrification.
err ...clarification ... I going to bed ... heh heh.
Zaux...from what i get is
there's 5 kids
take any 2 at random and weight them
do this 10 times and you get the paired weights.
hi Knightmare ... that interpretation works ... thanks
Pequod... er, I mean, Knightmare's on form tonight.
have the logic figured out but
something is wrong (or maybe I have an error)
I chose kids named a to e as lightest to heaviest. I have the weights except for b, one equation says b should be one more pound and another says one pound less.
Each kid gets weight 4 times --with each other kid.
The total weight is 1212 pounds divided by 4 gives 303 pounds.
Weights vary from 57 to 66 pounds.
Do you see an typing error
or a logic error I made?
e + d = 129
e + c = 125
e + b = 124
e + a = 123
d + c = 122
d + b = 121
d + a = 120
c + b = 118
c + a = 116
b + a = 114
e = 66
d = 129-e
c = 125-e
b = 124-e
a = 123-e
increasing e from 66 reduces the total weight from 303.
decreasing e from 66 increases the
total weight from 303
e 66
d 63
c 59
b 58
a 57
tot 303
Two of the equations
c + b = should be 118 but I get 117
b + a = should be 114 but I get 115
Telling me that theres and error somewhere.
Individual weights are 65, 64, 60, 58 and 56.
The two heaviest weigh 129 between them. The two lightest weigh 114 between them.
Go from there.
To elaborate, using Ragknot's notation:
e+d = 129 and e+c = 125, so d = c+4
The next highest pairs of weights are 124 and 123. These must be e+b and c+d.
Since d = c+4, c+d must be even and therefore 124.
So, d = 64 and c = 60, therefore e = 65.
At the other end of the scale, a+b = 114 and a+c = 116.
Since c = 60, a must be 56, hence b = 58.
Wiz
using your numbers, the top and bottom are ok, but the middle
ones are off.
TRUE 129 e + d = 129
TRUE 125 e + c = 125
FALSE 123 e + b = 124
FALSE 121 e + a = 123
FALSE 124 d + c = 122
FALSE 122 d + b = 121
TRUE 120 d + a = 120
TRUE 118 c + b = 118
TRUE 116 c + a = 116
TRUE 114 b + a = 114
My table needs explaining.
The first line says True d+e does equal 129.
This is what I gathered from your comment
e 65
d 64
c 60
b 58
a 56
The values I had have only two
Falses, and the counter each other
about the correction
e 66
d 63
c 59
b 58
a 57
TRUE 129 e + d = 129
TRUE 125 e + c = 125
TRUE 124 e + b = 124
TRUE 123 e + a = 123
TRUE 122 d + c = 122
TRUE 121 d + b = 121
TRUE 120 d + a = 120
FALSE 117 c + b = 118
TRUE 116 c + a = 116
FALSE 115 b + a = 114
Logic:
E is weighed 4 times, once each with a,b,c, and d.
Since their weight descends from d to a, then E+d = 129. 125 the next must be e and c. D can not be heavier than E with any kid, or the A to C would not be increasing.
Following this logic
e + d = 129
e + c = 125
e + b = 124
e + a = 123
d + c = 122
d + b = 121
d + a = 120
c + b = 118
c + a = 116
b + a = 114
Since the top four totals has E with each kid, you can assume a value for E and compute D thru A. Then you can verify the other totals, but the totals of 118 and 114 show something is wrong.
I think the trick is that B either went to the bathroom or cafeteria and lost or gained a pound. We don't know the weighing order, we don't know which, but sometime in the ten weighing his weight changed.
Ragknot,
The problem with your logic is, if you call the weights a through e lightest to heaviest then you can only be certain that:
The heaviest weight is: d+e
The second heaviest is: c+e
The lightest weight is:a+b
The second lightest is:a+c
Beyond that there are different scenarios where each increasing weight could belong to a different pair. Your attempt has preassigned those pairs, which may or may not be correct for each.
If I said much more than that, I'd just post my analysis.
Cam
Thanks Cam,
If change my criteria, the only solution that works is
E= 65
D= 64
C= 60
B= 58
A= 56
TRUE 129 e + d = 129
TRUE 125 e + c = 125
TRUE 124 d + c = 124
TRUE 123 e + b = 123
TRUE 122 d + b = 122
TRUE 121 e + a = 121
TRUE 120 d + a = 120
TRUE 118 c + b = 118
TRUE 116 c + a = 116
TRUE 114 b + a = 114
so Cam, you must be right.
The A to E is still increasing, but just not quite like I had thought
I did allow for what Cam just said about assigning weights to pairs (see my Jan 18 12.53am post above), but I did make one assumption that I should have validated.
(If anyone is still interested . . .)
I assumed that the 3rd and 4th highest weights (124 and 123) had to be between e+b and c+d and, because c+d is even, that had to be 124.
However, I overlooked the possibility that the 124 and 123could in fact be e+b and e+a. But this would lead to b = a+1. Since b+a, the lowest pair of weights, is 114, which is even (and we're assuming whole number weights for each kid) this can't be so.
So, my analysis, and the answers I gave above, still stand. I claim the non-existent prize!
Using Ragknot's notation have: e > d > c > b > a
Note no two girls can have the same weight, otherwise there would
be duplicate entries in the weighing table.
I'm pretty sure that if the girls weights weren't integers, the
problem would not have a unique answer.
Write out the table of weighings in the neatest order:
e+d, e+c, e+b, e+a, d+c, d+b, d+a, c+b, c+a, b+a
We can only be sure that e+d is the heaviest pair, and that a+b
is the lightest pair.
The sum of the weighings is 1212. Each girl has been weighed
four times, so the total weight of the girls is 1212/4 = 303.
But e+d = 129 so a+b+c = 303-129 = 174 and a+b = 114 so c = 60.
If you do a check, you'll easily find that if there are n girls
whose weights are odd, then for n = 1, 4 or 5 you'd get 4 odd
weighings, and for n = 2 or 3 you'd get 6 odd weighings.
So there are 1, 4 or 5 girls with odd weights (as we have four
odd weighings). But one of e or d is even, so n <> 5 and so
n = 1 or 4. So a, b and c are all even or all odd. But c is
even, so a and b must be even as well.
c > b > a. (a+b)/2 = 57 and both a and b are even. The only
possibility is that a = 56 and b = 58 => a+c = 56+60 = 116 and
that is seen to correspond to the second lightest weighing.
Now b+c = 58+60 = 118 and that is seen to be the third lightest
weighing. i.e. the three lightest weighings completely use up
the a, b and c combinations.
The fourth lightest weighing must involve d. And hence is a+d.
But a+d = 120 => d = 64. Then e = 129 - 64 = 65. We're done.
If it had turned out that b+c = 120, then I'm pretty sure that
I would have been able to proceed logically. The nearest I got to
trial and error was in obtaining the a and b values. But there
again the first trial was all that was needed.
Wiz, I think you've got a circular argument involved when you
assert that c+d is even. i.e. you guessed the answer then later
used the guess as if it were a logical deduction. I may retract
after studying it better with a coffee under the belt.
The site I got it from, completely glossed over the answer, as
it does with nearly all the answers to the more difficult
problems. It claims to be an educational site - I can't agree.
Hi Zuax, you dno't need to psot a ceroctoin for erevy tpyo. We all do it in the haet of the moenmt. It's olny wrtoh wihle if it cuseas mninaeg to be lsot.
Chris,
I hope that coffee has sharpened your mind!
There was no circular argument whatsoever in my proof. If you follow it through (including the last bit just before your latest post) you will see that I have clearly demonstrated that d = c+4. So d+c has to be even. The rest follows.
No trial and error is involved in any part of my proof. All logic! Simpler than yours, too!
I look forward to your promised retraction!
Another point: you say that you can only be sure that e+d is the heaviest pair and a+b the lightest. In fact, if you think about it, e+c must be the second heaviest and a+c the second lightest.
Hi Wiz, I ended up posting on the 100 problem. You're right about the second heaviest and lightest. I mucked that up during the many revisions of my post before posting it.
I hereby officially retract my (tentative) disparagement of your work and also admit that your answer is better than mine.
My solution suffered because having found a way to decide how many odd values there were, I hung on to it like a dog with a bone.
Wiz, nice answer and greetz.
Chris,
Official retraction officially accepted.
Forgive my vigorous defence of my efforts - it's not often I get one right!
Cheers.
Kris ... eye c whut ya mene! but whut duz it hert? dint relize i nedid specul permishun to tipe wun
oops ...Zaux
Doubling up:
Given 10 weighings
Call weights of 5 from lightest to heaviest:
x, x+a, x+b, x+c, x+d
Lightest weighing= x+(x+a)=2x+a
2nd lightest weighing: 2x+b
Heaviest weighing: 2x+c+d
2nd Heaviest weighing: 2x+b+d
2nd lightest weighing -2nd heaviest weighing:
(2x+b+d)-(2x+b)=d
d=125-116=9
For 3rd lightest weighing:
It could be 2x+c OR if (a+b)<c , 2x+a+b
Assume it is 2x+c
Heaviest weighing- 3rd Lightest weighing
(2x+c+d)-(2x+c)=d
129-118=11 but d=9, so 2x+c is invalid, it must be 2x+a+b
2nd lightest weighing-lightest weighing
(2x+b)-(2x+a)=b-a
116-114=2=b-a
b=a+2
3rd lightest weighing-Lightest weighing:
3rd lightest weighing is 2x+a+b=2x+2a+2
(2x+2a+2)-(2x+a)=a+2
118-114=4
4=a+2
a=2
Lightest weighing=2x+a=2x+2=114
x=56
x+a=56+2=58
x+b=56+2+2=60
Heaviest weighing = 2x+c+d
129=2*56+c+9
c=8
x+c=56+8=64
x+d=56+9=65
Answer:
The individual weights are:
56,68,60,64,65
Cam
Wiz, I should have known better than to challenge you on logic.
Hi Cam. That was good. I liked the sleight of hand with x. Thanks for posting it. It also is better than my effort.
I hope you agree that Wiz has the best solution though.
Zaux. Of course you don't need [my] permission to post. The main purpose of my post was to amuse, not admonish. Your unnecessary typo corrections simply reminded me of how we can read pretty well if only the first and last letters of the words are in the right place. Silly thing is, you didn''t do it on this particular blog.
129, 125, 124, 123, 122,121, 120, 118, 116 and 114
e > d > c > b > a
e+d must be the heaviest and e+c the second heaviest weighing.
a+b must be the lightest and a+c the second lightest weighing.
Sum of weighings is 1212 => a+b+c+d+e = 303
c = (a+b+c+d+e) - (a+b) - (d+e) = 303 - 114 - 129 = 60
a+c = 116 => a = 116 - 60 = 56
a+b = 114 => b = 114 - 56 = 58
c+e = 125 => e = 125 - 60 = 65
e+d = 129 => d = 129 - 65 = 64
So have 56, 58, 60, 64 and 65
I note that the above didn't make an appeal to the values being integers and that the answer is unique.
I declare myself the winner :)
I declare you the winner, too, Chris.
Definitely the best proof.
Although it Was Wiz who reminded me about the penultimate weights, being e+c and a+c, I acknowledge tha Cam had also published that, very clearly.
Here's my "logic". It took 0.005 seconds to run. But it took several hours and Cam's help to correct the "Test" subroutine near the end.
Sub weightin()
'OUTPUT
'e = 65
'd = 64
'c = 60
'b = 58
'a = 56
'1 Solution(s)Found.. 0.005 seconds
'Five schoolchildren were weighed in pairs.
' Their (combined) weighs were:
'129, 125, 124, 123, 122,
'121, 120, 118, 116 and 114 lbs.
'What were their individual weights?
time1 = Timer * 100
tot = (129 + 125 + 124 + 123 + 122 + 121 + 120 + 118 + 116 + 114) / 4
For a = 50 To 70: b = 114 - a
For c = 50 To 70
For e = 50 To 70: d = 129 - e
If a + b + c + d + e = tot Then
GoSub test
End If
Next e
Next c
Next a
GoTo Ennd
test:
If e + d <> 129 Then Return
If e + c <> 125 Then Return
'I had other tests here, but Cam showed they might not be correct. I found they weren't needed - (Thanks Cam)
If c + a <> 116 Then Return
If b + a <> 114 Then Return
Debug.Print "E="; e
Debug.Print "D="; d
Debug.Print "C="; c
Debug.Print "B="; b
Debug.Print "A="; a
i = i + 1
Return
Ennd:
Debug.Print i; "Solution(s) Found"
time2 = Timer * 100
Debug.Print (time2 - time1) / 100; " seconds"
End Sub
(Reposted due to an unfortunate ambiguity).
Hi Ragknot, thanks for the code. A nice thing about your method is that it shouldn't be difficult to extend to harder cases: I expect that the analytic methods would start to get quite messy with 6 or more girls involved. Also, it might not be that hard to tweak it to work with non-integral weights, and errors in the weighings. It would also deal with sets of girls with the same weight.
I think that the completely different approach to solving problems when computers are involved is quite amazing.
For the record, Wiz was the first to post. He also gave a clear, straightforward ananlysis.
When I finally posted the most straightforward analysis (possible!), I was pretty sure that Cam and Wiz would laugh at themselves for having missed it themselves (I did). I declared myself the winner both for a joke and as a serious assesment of the quality of the answer.
Ragknot's programmatic method (after suitable tweaking) would probably be the only practical way of solving such a problem if more girls were involved.
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