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

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

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

Bunuel · Accepted Answer

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.

mbaiseasy · Answer

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.

shrouded1 · Answer

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)

onedayill · Answer

Both the solution mentioned above are good,

Is there any quicker way to deal with this.

shrouded1 · Answer

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

adishail · Answer

Bunuel's method is faster, but needs slightly more undertstanding.

GSBae · Answer

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.

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.

aj33 · Answer

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

HouseStark · Answer

TOTAL NO. OF WEIGHT this 6 {1Kg, 2Kg, 4Kg, 8Kg, 16Kg, 32Kg} CAN MEASURED= 1KG TO 63KG=  TOTAL 63

IF 4KG WEIGHT IS LOST

1KG______2KG_____8KG_______16KG_____32KG

63-(1+\frac{5!}{4!}+\frac{5!}{2!*3!}+\frac{5!}{2!*3!}+\frac{5!}{4!})=32

ANSWER D

Albymana · Answer

My view on this problem

Sum may be seen as a result of selections of 1,2,3,4,5 elements

When weights are 6 you can have

6C1 + 6C2 +.... + 6C6= 63

After losing one you can have

5C1 + 5C2 + .... 5C5 = 31

Differences between 63 and 31 may be seen as the selection you can no more have

Posted from my mobile device

riturajsingh21 · Answer

I Counted from 1 to 63. Numbers you can not count without 4 K.G. Weight :

4
5
6
7

12
13
14
15

20
21
22
23

etc..

4*8=32

Yosemite98 · Answer

the easiest way to me wihtout knowing any theorem is the following logic.

I will count how many weights I can make with a 4KG and the other weights.

1. I only sum 2 elemets -> 4 + X -> Total posibilities 5
2. I sum 3 elemets -> 4 + X + Y -> Total possibilities = 5!/(3!2!) = 10
3. I sum 4 elemets -> 4 + X + Y + Z -> Total possibilities = 5!/(2!3!) = 10
4. I sum 5 elemements -> 4 + + X + Y + Z + W -> Total possibilities = 5!/(1!4!) = 5
 5. I sum 6 elemements -> 4 + + X + Y + Z + W + M -> Total possibilities = 5!/(5!) = 1

Total = 31 so the new possiblities are 63-31 = 32

aryanxx09 · Answer

Can we not assume that he can use other weights to keep on the other side and get the actual weight by balancing that out?

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,

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