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BMAD

if a,b,c, and d are natural numbers revisited.

Question

Suppose there is a rock, R.  It's weight is known to be 40 lbs.  The store owner uses the rock to measure out purchases; in other words, customers bought 40 lbs of an item when making a purchase.  He would load the rock on one end of a scale and the food on the other until they are even then the transaction occurs.  One day, the clerk drops the rock and it breaks into exactly four distinct pieces (e.g. R= A+B=C+D),  At first he thought he was in trouble as he may not have any way to sell his items.  But fortunately he found that he can still stack all 40 lbs on the same side of the scale to complete a 40 lb purchase.  Additionally, he found that he can now use the various pieces on different sides of the scale to allow for a customer to buy products anywhere from 1 to 40 lbs ultimately bringing in even more customers.

If A>B>C>D >0 

prove A-B-C-D > C-D

bonus points if you can explain (warning gives the answer away)...

Why do the elements in the set R follow the pattern of Rn = 3, where Rn represents the elements in R and A = R0

Edited by BMAD
bonus question.

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8 answers to this question

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Just to be clear about the order of subtraction.

In your inequality, does A - B - C - D  =   A - (B + C + D)?

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Just to be clear about the order of subtraction.

In your inequality, does A - B - C - D  =   A - (B + C + D)?

yes but i did find a typo. i meant R=A+B+C+D

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Anyone going to attempt the bonus? It extends with R have any number of elements and being able to 'make' any number.

 

Hidden Content

nice!

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Proof for part 2

"Why do the elements in the set R follow the pattern of Rn=3n , where Rn represents the elements in R and A = R0"

Using induction: "Define Rn=3n. Any element between 1 and Sum(Rn) can be weighted using R0, ... Rn"

This holds for n=1. Having 1 and 3 as weights, one can measure weight 1 using weight 1 on the other side, weight 2 using 1 on the same side and 3 on the other (2=3-1), 3 using only 3 on the other side and 4 using both on one side.

Assume this holds for some n>=1. 

Denote Sum(Rn)=S and Sum(Rn+1)=T where sums are for all 0<=k<=n, and 0<=k<=n+1 respectively (a simplified sum notation showing only the last element).

It follows trivially that T-1=3*S. Next Rn+1 -1 = 3(n+1)-1(n+1)=(3-1)*(3n+...+3+1)=2*S.

There are 4 sets of products between 1 and T=3*S+1.

  • First third are those products between 1 and S, where weights R0, ... Rn would suffice by induction hypothesis.
  • Second third are products between S+1 and 2*S. Seeing as Rn+1=2*S+1, if we put Rn+1 on one scale and the product on the other, the scale tips to Rn+1 tips towards Rn+1 which is bigger and the difference is between 2*S+1-(S+1)=S and 2*S+1-2*S=1. But we can balance this difference (since it's between 1 and S) considering it equivalent to the scenario having a single product of that weight instead and using weights R0, ... Rn by induction hypothesis.
  • Last third are products between 2*S+1 and 3*S+1. Seeing as Rn+1=2*S+1, if we put Rn+1 on one scale and the product on the other, it either balances perfectly or it tips towards the product with a difference between 3*S+1-(2*S+1)=S and 2*S+1-(2*S+1)=0. But we can balance this difference (since it's between 0 and S) considering it equivalent to the scenario having a single product of that weight instead and using weights R0, ... Rn by induction hypothesis.

This only shows that the proposed set has the property for all n>0.

Conversely, it is the set that yields the maximum product weight Sum(Rnthat can be weighted using the set of weights with minimal sum. The proof is less simple since one has to show that the minimal set that can be used to weight everything from 1 to Sum(Rn) has at the same time two properties - it has exactly n+1 weights and the sum is Sum(Rn). I am 100% positive, but I lack the clarity of proof. Or in other words, the margin is too narrow for the proof :P

 

 

 

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