# Weighing the Merchandise !

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A Hawker sets out to sell his wares in the Market !

In those British Rule days in India Pounds [ Lbs ] are used to weigh the merchandise.

The Hawker has a Simple Balance which he uses to weigh his merchandise to sell to his Customers..

But he has only one Measure of Weight which weighs 40 Pounds !

He needed Units of weight with which he can measure out 1, 2, 3 etc upto 40 Lbs in one weighing !

Of Course he can put a weight or pieces of weight in one Pan while the other Pan can contain the merchandise and weights if needed.

If he can CUT THE WEIGHT into 40 equal parts each weighing 1 Lb he can weigh out upto 40 pounds in one weighing but then it will be a laborious process.

He asked me to cut the 40 Lbs weight into minimum number weights with which he can handle 1,2,3,4,5 etc upto 40 Lbs in one weighing !

How many pieces I had to cut the 40 Lbs weight into and what are the weights of each Piece and how can he weigh out 1,2,3 etc upto 40 Pounds of merchandise in one weighing using those Pieces of Weight ?

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I've just guessed at 20-10-6-3-1 which seems to give any weight up to 40 (but it's Monday morning)

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Edit: It's Monday for me too. Post deleted.

Edited by zapatos

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I've just guessed at 20-10-6-3-1 which seems to give any weight up to 40 (but it's Monday morning)

Yes that's a nice answer ! Anything better ?

+1 to you. But I am waiting for the Right answer !

Edited by Commander

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I thought this will be quite a simple puzzle to solve !

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Sorry, I can't improve on my previous stab in the dark.

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You can do it in 4 weights not 5.

Cut your 40 into a 1, 3, 9, 27. In the below list a positive number is a weight in the normal pan whereas a negitive number is a weight in the merchandise pan - a zero is the weight not being used

0 ,0 ,0 ,0 = 0
1 ,0 ,0 ,0 = 1
-1 ,3 ,0 ,0 = 2
0 ,3 ,0 ,0 = 3
1 ,3 ,0 ,0 = 4
-1 ,-3 ,9 ,0 = 5
0 ,-3 ,9 ,0 = 6
1 ,-3 ,9 ,0 = 7
-1 ,0 ,9 ,0 = 8
0 ,0 ,9 ,0 = 9
1 ,0 ,9 ,0 = 10
-1 ,3 ,9 ,0 = 11
0 ,3 ,9 ,0 = 12
1 ,3 ,9 ,0 = 13
-1 ,-3 ,-9 ,27 = 14
0 ,-3 ,-9 ,27 = 15
1 ,-3 ,-9 ,27 = 16
-1 ,0 ,-9 ,27 = 17
0 ,0 ,-9 ,27 = 18
1 ,0 ,-9 ,27 = 19
-1 ,3 ,-9 ,27 = 20
0 ,3 ,-9 ,27 = 21
1 ,3 ,-9 ,27 = 22
-1 ,-3 ,0 ,27 = 23
0 ,-3 ,0 ,27 = 24
1 ,-3 ,0 ,27 = 25
-1 ,0 ,0 ,27 = 26
0 ,0 ,0 ,27 = 27
1 ,0 ,0 ,27 = 28
-1 ,3 ,0 ,27 = 29
0 ,3 ,0 ,27 = 30
1 ,3 ,0 ,27 = 31
-1 ,-3 ,9 ,27 = 32
0 ,-3 ,9 ,27 = 33
1 ,-3 ,9 ,27 = 34
-1 ,0 ,9 ,27 = 35
0 ,0 ,9 ,27 = 36
1 ,0 ,9 ,27 = 37
-1 ,3 ,9 ,27 = 38
0 ,3 ,9 ,27 = 39
1 ,3 ,9 ,27 = 40

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You can do it in 4 weights not 5.

Cut your 40 into a 1, 3, 9, 27. In the below list a positive number is a weight in the normal pan whereas a negitive number is a weight in the merchandise pan - a zero is the weight not being used

0 ,0 ,0 ,0 = 0

1 ,0 ,0 ,0 = 1

-1 ,3 ,0 ,0 = 2

0 ,3 ,0 ,0 = 3

1 ,3 ,0 ,0 = 4

-1 ,-3 ,9 ,0 = 5

0 ,-3 ,9 ,0 = 6

1 ,-3 ,9 ,0 = 7

-1 ,0 ,9 ,0 = 8

0 ,0 ,9 ,0 = 9

1 ,0 ,9 ,0 = 10

-1 ,3 ,9 ,0 = 11

0 ,3 ,9 ,0 = 12

1 ,3 ,9 ,0 = 13

-1 ,-3 ,-9 ,27 = 14

0 ,-3 ,-9 ,27 = 15

1 ,-3 ,-9 ,27 = 16

-1 ,0 ,-9 ,27 = 17

0 ,0 ,-9 ,27 = 18

1 ,0 ,-9 ,27 = 19

-1 ,3 ,-9 ,27 = 20

0 ,3 ,-9 ,27 = 21

1 ,3 ,-9 ,27 = 22

-1 ,-3 ,0 ,27 = 23

0 ,-3 ,0 ,27 = 24

1 ,-3 ,0 ,27 = 25

-1 ,0 ,0 ,27 = 26

0 ,0 ,0 ,27 = 27

1 ,0 ,0 ,27 = 28

-1 ,3 ,0 ,27 = 29

0 ,3 ,0 ,27 = 30

1 ,3 ,0 ,27 = 31

-1 ,-3 ,9 ,27 = 32

0 ,-3 ,9 ,27 = 33

1 ,-3 ,9 ,27 = 34

-1 ,0 ,9 ,27 = 35

0 ,0 ,9 ,27 = 36

1 ,0 ,9 ,27 = 37

-1 ,3 ,9 ,27 = 38

0 ,3 ,9 ,27 = 39

1 ,3 ,9 ,27 = 40

imatfaal : Absolutely Right & that is the answer I was looking for and the sequence is 30 , 31 , 32 , 33 for OPTIMUM QUANTIFICATION with negative Symbol Inclusion !

Well done +1 to you