**Weighing Balance Puzzle - 8 September**

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

**For Solution :**Click Here

Total 7 weight of 1, 3, 9, 27, 81, 243, 729

ReplyDelete10 weights all should power of 2

ReplyDeleteit will be a GP series...7 weights...1, 3, 9, 27, 81, 243, 729

ReplyDeleteTotal weights required will be be 7, these are 1,3 ,9, 27, 81,243,729

ReplyDelete@rizwan & vishwas then tell how to measure 38...

ReplyDelete27, 9, and 3 (39 kg) on one side, 1 on the other.

Delete38 is simply 3, 9 and 27 on one side and 1 on the side with the substance to be weighed.

ReplyDelete(i.e. 3+9+27-1=38)

So By this is there a way to measure 20 ?....

ReplyDeletePlate A: Substance, 9, 1

DeletePlate B: 27, 3

27+3-9-1 = 20

yes..(27+3)-(9+1)

ReplyDeleteIts simple bhai log..1,2,4,8,16,32,64,128,...512

ReplyDeleteIt is simply 1,3,9,27,81,243,636.the last denomination won't be 729 as you can see that sum of all denomination should be 1000..:)

ReplyDelete1000 = 729 + 243 + 27 + 1

Deleteyou can either agree with us or be wrong

how can u measure 7?

Delete7 can be measured as (9+1) - 3

Delete10 are required.... since ceiling(log2(1000)) is 10

ReplyDeleteonly 1kg is sufficient :-p

ReplyDeleteThis comment has been removed by the author.

ReplyDelete@Dhayanithi : 7 can be measured as (9+1) - 3

ReplyDelete10 will be required of denominations 1,2,4,8,16,32,64,128,256,512

ReplyDeletelook above... yours works but fewer weights (the goal of the problem) is possible. your method allows multiple ways to get the same result (for example: (8-1)=(4+2+1)). 1 3 9 27 81 243 729 is the simplest possible solution to explain. there are others that work and are also 7 weights but 7 is the minimum.

DeleteHow can u solve this plz tell me trick to solve such type of question

ReplyDeleteYes. Can anybody tell how you came up with this solution??

ReplyDeleteBoth power of 2 and power of 3 works but power of 3 works with less number of weights.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThe numbers 1,2,4,8,16... This is optimal as each number has exactly one binary representation. So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, 512. :)

ReplyDeleteThe powers of 3 solution is correct. The way this is derived is as follows:

ReplyDeleteWe have n weights and k possibilities of how to combine them. For each weight, we can choose it to be positive, negative, or zero. Thus, with n=1, we have k=3.

Let us call each weight by a different letter (a,b,c etc.)with each weight being greater than the last. Thus, with n=1 our possibilities are +a, 0, -a. Ignore negative values for now, I will come back to that.

We now examine the case of n=2. It should be clear that there are 9 possibilities. For each value of a (+,0,-) we attach a value of b (+,0,-) such that there are 9 total values (-a,0,a,b-a,b,b+a,-b-a,-b,-b+a). Again ignore the negatives.

We see now that when n is incremented, k[n] = k[n-1]*3. Or, k[n]=3^n.

We now must examine which of our values are positive. For each new n, the positive values are those that were previously positive to which we add 0, and all previous values to which we add the new weight. Verify this for the n=2 case: new positive values are the previously positive values (a) and all previous values to which we add b (b-a, b, b+a). It can be seen that there are Sum[i=1,n] (3^(n-i)) positive values. We will call these positive values j.

It is now a simple matter to solve for 1000<= j. This gives us n=7. This means we must have 7 weights. It is simple to verify from here that each weight must be of value k. Thus, we have weights of k^0, k^1 ... k^6, or 1,3,9,27,81,243,720

awesome

Delete1,2,3,4,20,30,40,200,300,400

ReplyDeletecould any one will help me to understand in better way????

ReplyDeleteYes! Two steps.

Delete1 For a weight(thing you use to measure other items), to weight an object, you have 3 ways to put it on the balance, you put it together with the object you want to weight on one side, or you do not use it, or you put it on other side of the balance. e,g, you have a weight labelled with w1, when you weight an object x, you could put w1 with x on left of the balance, or do not use w1, or put it on the right of balance. So, after measured it, The weight of right balance is 100kg; for 3 above-mentioned cases, the weight of object x will be 100-w1, 100, 100+w1.

So ideally, you would have 3 different ways for each weight, and get 3 (not necessary unique) measures. Hence, theoretically you need log3(1000) weights. That will be 6.x. That will be at least 7.

2 Step, 7 will be a theoretical best result. You need to check or prove it is really possible since you might have reptitions of measures from different weights combination. That's how Geometric Progression series come into play. 1, 2, 4, 8 could do it, but it is too costly. We need to try out my luck on 1,3,9,27,...

1=1

2=3-1

3=3

4=3+1,

5=9-3-1

6=9-3

7=9-3+1

8=9-1

...

Wow, what a beautiful GP series. You strike a jackpot, it is just the ideal case. No thing could be better than this. This could be simply proved by math induction.

Summary: If you are interested, go take some number theory lecture, you will be sure to find some amazing "thing" about those numbers. The most famous one will be Goldbach Conjecture ("Every even integer greater than 2 can be expressed as the sum of two primes")

This comment has been removed by the author.

ReplyDeleteI have come up with an alternate explanation for the benefit of people who wanted explanation below:

ReplyDeleteLet us say we have already found n-1 items in the series and have to find the nth one. The n-1 items would have already ensured everything from 1 to Sigma [k(n-1)] can be achieved through various combinations (because the maximum number that can be derived from k(1), k(2)... k(n-1) is Sigma [k(n-1)] ). Now if you choose 2*Sigma[k(n-1)]+1 as the nth item in the series, you are getting the maximum benefit as every number between Sigma [k(n-1)] and 2*Sigma[k(n-1)]+1 can be expressed in the form of 2*Sigma[k(n-1)]+1 - a where a is a number <= Sigma [k(n-1)]. So this gives the series equation as below:

k(n) = 2*Sigma[k(n-1)]+1 ------- (1)

So, k(n-1) = 2*Sigma[k(n-2)]+1 ---------- (2)

expanding equation (1) above,

k(n) = 2*{k(n-1) + Sigma[k(n-2)]} + 1 ---------- (3)

substituting equation (2) in equation (3),

k(n) = 2*{2*Sigma[k(n-2)]+1 + Sigma[k(n-2)]} + 1

k(n) = 2*{3*Sigma[k(n-2)]+1} + 1

k(n) = 6*Sigma[k(n-2)] + 2 + 1

k(n) = 6*Sigma[k(n-2)] + 3

k(n) = 3*{2*Sigma[k(n-2)]+1} ---------- (4)

Using equation (2) in equation (4) above,

k(n) = 3*k(n-1) ------------ (5)

Equation (5) above shows this is a series of powers of 3.

Hence, the answer will be 1, 3, 9, 27, 81, 243, 729

how about my answer?

ReplyDelete600/300/100

60 /30 /10

6 /3 /1

I know my answer is funny.!! It need 6 weights for 777kg. :D

You guys are missing the spirit of the questions. You missed the phrase 'For example,' the weights were never defined explicitly. You only need 1 weight to determine 1000Kg in weight, a 1000Kg weight.

ReplyDeleteHi guys, my answer is 1, 2, 3, 4, 20, 50, 100, 200, 500, 1000. Here, it is easy to sum and minus and get any weight in less time( within seconds) in practically. These 1,3 ,9, 27, 81,243,729 and 1,2,4,8,16,32,64,128,256,512 are also right for getting less numbers of weights to be needed to weigh between 1 to 1000. But it is difficult to sum and minus to get required weight and it takes more time. Theoretically it is easy to say but practically it is difficult to use.

ReplyDeleteit's 7, powers of three so 1, 3, 9, 27, 81, 243, 729.

ReplyDeleteExplanation:

Start with 1, which can only make 1 (duh). Double it so you get 2, then add 1, giving 3. Why? because this means you can 1, 2, 3 and 4 by doing 1, (3-1), 3 & (3+1)

Double the 4 and add 1 to get 9. Then you can still make 1,2,3 & 4, but can also subtract all for these from 9 to make 5,6,7 & 8, make 9, or add them on to make 10,11,12 or 13.

Repeat process again and again to get the other numbers. This is power three numbers because 1 is 3^0. n=1

the next would be:2n+1

then:(2*((2n+1)+n)+1)=6n+3

then:(2*((6n+3)+(2n+1)+n)+1)=18n+3

and so on, with each value being three times the previous. if the first value is 1, then all of the values are powers of three.

whoops, the third equation is 18n+9

DeleteHi I guess all you guys did not understand the question. The question specifically ask to measure all weights between 1 to 1000.

ReplyDeleteExample you want to weigh 427....now your selections should give you this result.

My solution:

I need 3 kg weights ( 3 units) and 1 kg weight (1 unit) to measure anything between 1 and 10 kg.

Than I will have 10 kg weights ( 9 units) to measure anything between 1-100 kg.

example 76 kg. ( 10x7 + 3x2 )

Next I will have 100 kg weights ( 9 in numbers) to weigh anything from 101-1000 kg.

Total No of units = 3 + 1+ 9+9 = 22

Units by weight:

1 kg : 1 unit

3 kg : 3 unit

10 kg : 9 unit

100 kg: 9 unit

I hope this is satisfactory.

1000 x a 1kg weight.

ReplyDeleteThe question is : how many.... minimum weights... do you need to measure all weights from 1kg to 1000kg.

The only way to get 1000kg with minimum weights is 1000 since 1kg is the minimum weight.

The question should have been. What is the minimum amount of weights do you need to weight all weights from 1kg to 1000kg.