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A scientist has a set of weights {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, [#permalink]
01 Oct 2010, 10:19

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60% (02:24) wrong based on 47 sessions

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 ?

Re: Missing Weights [#permalink]
01 Oct 2010, 10:54

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Expert's post

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.

Re: Missing Weights [#permalink]
01 Oct 2010, 11:02

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

Re: Missing Weights [#permalink]
01 Oct 2010, 16:37

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 _________________

Re: Missing Weights [#permalink]
01 Oct 2010, 19:19

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. _________________

Re: Missing Weights [#permalink]
20 Dec 2012, 21:31

2

This post received KUDOS

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. _________________

Impossible is nothing to God.

gmatclubot

Re: Missing Weights
[#permalink]
20 Dec 2012, 21:31

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