Suggest-minimum-number-of-weights-by-which-we-can-weigh-upto-40-kgs-in-whole-numbers-in-a-traditional-balance-If-possible-mention-the-process-also-

Question Number 3143 by Rasheed Soomro last updated on 06/Dec/15

S u g g e s t m i n i m u m n u m b e r o f w e i g h t s b y w h i c h w e c a n

w e i g h u p t o 40 k g s (i n w h o l e n u m b e r s) i n a t r a d i t i o n a l b a l a n c e .

I f p o s s i b l e m e n t i o n t h e p r o c e s s a l s o .

Commented by Rasheed Soomro last updated on 06/Dec/15

I n a t r a d i t i o n a l b a l a n c e w e c a n p l a c e w e i g h t s o n b o t h

s i d e s .

Answered by prakash jain last updated on 06/Dec/15

You start with 1 kg

For 2 kg select 1 + 2 = 3 kg - you get 1 to 4

for 5 kg select 4 + 5 = 9 kg you get 1 to 13 kgs

for 14 kg select 13 + 14 = 27 kg you get 1 to 40 .

so you need following weight

1, 3, 9, 27 to weight all upto 40 kg (in whole numbers) .

Commented by RasheedAhmad last updated on 06/Dec/15

\overset{V}{N} i c e S i r!

Commented by RasheedAhmad last updated on 06/Dec/15

^{R a s h e e d S o o m r o}

A l l a r e p o w e r s o f 3 .

N e x t p o w e r i s 81

F i v e w e i g h t s (1, 3, 9, 27, 81) c a n

w e i g h u p t o 1 + 3 + 9 + 27 + 81 = 121 kg .

n w e i g h t s 3^{0}, 3^{1}, 3^{2}, \dots 3^{n - 1} c a n w e i g h

u p t o 3^{0} + 3^{1} + 3^{2} + \dots + 3^{n - 1}

= \frac{3^{n} - 1}{2}

Commented by prakash jain last updated on 06/Dec/15

Yes .

Commented by Yozzi last updated on 06/Dec/15

W h y i s y o u r a n s w e r c o r r e c t ? W h y

s t a r t w i t h 1 k g, t h e n 2 k g, \dots ?

I {d o n}^{'} t u n d e r s t a n d y o u r s t e p s .

(n o t d o u b t i n g {i t}^{'} s w r o n g; I w a n t t o

u n d e r s t a n d t h e i d e a b e h i n d t h e

q u e s t i o n .)

Commented by prakash jain last updated on 06/Dec/15

With 1 kg you can weigh up to 1 kg .

To weigh 2 kg the maximum weight

that we can use is 3 kg . This gives the maximum

range to 4 kg . 1, 3 - 1, 3, 3 + 1 .

Similarly we use the largest possible

weight every time to get the \max range so

that minimum number of weights are used .

Since we are able to weigh 1 to 4 (with 1 and 3)

adding 9 kg will allow addition range

(9 - 4) to (9 + 4) .

and so on .

1 = 1

2 = 3 - 1

3 = 3

4 = 3 + 1

5 = 9 - 4 = 9 - (3 + 1)

6 = 9 - 3

7 = 9 - 2 = 9 - (3 - 1) = 9 + 1 - 3

8 = 9 - 1

9 = 9

10 = 9 + 1

11 = 9 + 3 - 1

12 = 9 + 3

13 = 9 + 3 + 1

When you add 27 you can get 27 - (1 to 13) to

27 + (1 to 13) . Since 1 to 13 can be contructed

without using 27 kg weight .

Since we can put weights on both sides of

balancr subtraction is allowed .

Commented by prakash jain last updated on 07/Dec/15

Suppose you have n weight . The possible

weights that you can create (including

- ve and duplicates) are .

choose 1 and put a + / - sign before it = 2^{n} C_{1}

choose 2 and put a + / - sign before each = 2^{2}^{n} C_{2}

and so on

So total number of sum that you can

obtain with n + v e numbers are = \underset{i = 1}{\sum^{n}} 2^{i}^{n} C_{i}

{(1 + 2)}^{n} =^{n} C_{0} + 2 \cdot^{n} C_{1} + 2^{2}^{n} C_{2} + \dots + 2^{n}^{n} C_{n}

3^{n} = 1 + \underset{i = 1}{\sum^{n}} 2^{i}^{n} C_{i}

\underset{i = 1}{\sum^{n}} 2^{i}^{n} C_{i} = 3^{n} - 1

w i t h 3 + ve n u m b e r y o u c a n c r e a t e = 3^{3} - 1 = 26

half of them will be - ve so 13 + ve number .

with 4 + ve number you can create 3^{4} - 1 = 80

or 40 + ve numbers .

This proves that you need a minimum of

4 weights to create 40 + ve results .

Solving the question is about finding 4 numbers

so the sums are 1 to 40 .

Commented by Yozzi last updated on 06/Dec/15

A h, I n o w u n d e r s t a n d . {I t}^{'} s t h e t h o u g h t

b e h i n d i t t h a t I w a n t e d t o g r a s p .

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