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A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg,

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A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

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A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A) 16
B) 24
C) 28
D) 32
E) 36
[Reveal] Spoiler: OA

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Re: Missing Weights [#permalink]

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shrouded1 wrote:
Here is an interesting question :

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A) 16
B) 24
C) 28
D) 32
E) 36


Consider the following example: how many different selections are possible from \(n\) people (including a subset with 0 members and a subset with all \(n\) members)?

\(C^0_n+C^1_n+C^2_n+...+C^n_n=2^n\) --> so, number of different subsets from a set with \(n\) different terms is \(2^n\) (this include one empty subset). Or another way: each person has 2 choices, either to be included or not to be included in the subset, so # of total subsets is \(2^n\).

Next, from a set {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg} obviously no term can be obtained by adding any number of other terms.

So, from a set with 6 different terms {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg} we can form \(2^6-1=63\) subsets each of which will have different sum (minus one empty subset) --> we can weight 63 different weights;

From a set with 5 different terms {1Kg, 2Kg, 8Kg, 16Kg, 32Kg} we can form \(2^5-1=31\) subsets each of which will have different sum (minus one empty subset) --> we can weight 31 different weights;

Which means that if 4Kg weight is lost 63-31=32 weights can no longer be measured.

Answer: D.
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Re: Missing Weights [#permalink]

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New post 01 Oct 2010, 12:02
Good solution, alluding to a binary notation of a number.

Here is another way to think of the problem :

We can form :
1 {1}
2 {2}
3 {1,2}
But we cannot form any of 4 through to 7 without a 4.
So 4 numbers we cannot form up till 7

Next look at the next set of numbers from 8 to 15
Notice that this is set is nothing but :
8+{0,1,2,3,..,7}
But we know from this that we cannot form a 4,5,6,7
So if we count up till 15, there are 2x4=8 numbers in all that we cannot form

Next look at the numbers from 16 to 31
Same patter repeats
This is 16+{0,1,2,...,15}
And we know from 0,..,15 there are 8 numbers we cannot form.
So from 16 to 31 there are 8 more ... hence 2x8=16 numbers between 1 and 31

and this patter goes on ...

For numbers upto 63, it will be 2x16=32

Answer is (d)
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Re: Missing Weights [#permalink]

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New post 01 Oct 2010, 12:37
Both the solution mentioned above are good,

Is there any quicker way to deal with this.
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Re: Missing Weights [#permalink]

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New post 01 Oct 2010, 17:37
There is one more way to do this, but that involves knowledge of a binary notation of a number (a concept not tested on the GMAT). Here is the solution :

In binary a number <=63 can be represented in 6 digits.
Of this 4 represents the 3rd digit from the right.
The number of 6 digit binary numbers possible forcing the 3rd digit to be 1 (all the numbers that need 4) is exactly 2^5 or 32
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Re: Missing Weights [#permalink]

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New post 01 Oct 2010, 20:19
shrouded1 wrote:
Good solution, alluding to a binary notation of a number.

Here is another way to think of the problem :

We can form :
1 {1}
2 {2}
3 {1,2}
But we cannot form any of 4 through to 7 without a 4.
So 4 numbers we cannot form up till 7

Next look at the next set of numbers from 8 to 15
Notice that this is set is nothing but :
8+{0,1,2,3,..,7}
But we know from this that we cannot form a 4,5,6,7
So if we count up till 15, there are 2x4=8 numbers in all that we cannot form

Next look at the numbers from 16 to 31
Same patter repeats
This is 16+{0,1,2,...,15}
And we know from 0,..,15 there are 8 numbers we cannot form.
So from 16 to 31 there are 8 more ... hence 2x8=16 numbers between 1 and 31

and this patter goes on ...

For numbers upto 63, it will be 2x16=32

Answer is (d)


I was doing it this way ... however, lost trach somewhere in the middle. Realize that rather than doing all the way from 1 - 63 in one go, it is better to take it in batches. Thanks.
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Re: Missing Weights [#permalink]

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New post 01 Oct 2010, 20:20
onedayill wrote:
Both the solution mentioned above are good,

Is there any quicker way to deal with this.


Bunuel's method is faster, but needs slightly more undertstanding.
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Re: Missing Weights [#permalink]

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New post 20 Dec 2012, 22:31
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shrouded1 wrote:
Here is an interesting question :

A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg}. This set is good enough to weight any object having an integral weight betweem 1Kg and 63Kg (Eg. 19Kg = 16Kg + 2Kg + 1Kg). If the 4Kg weight is lost, how many weights between 1Kg & 63Kg can no longer be measured ?

A) 16
B) 24
C) 28
D) 32
E) 36


Originally there could be \(2*2*2*2*2*2=2^6=64\) combinations of weight. Now that we took 1 weight off, we get \(2*2*2*2*2=2^5=32\) combinations of weight.

What is lost? \(64-32=32\)

Answer: D

In case you are wondering why 2 was multiplied n times, it's because 2 represents two things: BEING SELECTED and NOT BEING SELECTED.
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Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

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Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

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New post 05 May 2016, 10:00
Here's how I thought about it: For each weight, there are two possibilities; either it's included or it's not. So, for 6 weights, there are (2*2*2*2*2*2) = 2^6 possibilities, minus 1 (the case where there are no weights) or 63 possibilities. Therefore, if we take away one of the weights, there are now 2^5 - 1 = 31 possibilities. 63-31 = 32.

Answer: D

Aside: Think about it similiarly to how you think about counting the total number of factors from a number's prime factorization: Either the factor is included or it's not.
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Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]

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New post 05 May 2016, 10:16
the first thought i had was 5c0 +5c1 + 5c2 + 5c3 +5c4 +5c5 = 32. This is because i need to choose a 4 and that can be selected in the following combinations. with each one. on solving it gives me 32. 32 sets will be lost.
Hope this helps
Re: A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg,   [#permalink] 05 May 2016, 10:16
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