They all look the same

Logic Level 1

I have 11 stacks of marbles, each stack consisting of 10 marbles. I am also given a digital scale with arbitrary precision. All of these marbles are indistinguishable.

Given that all the stacks of marbles are made of glass weighing 10 grams each except one stack which are all made from clay but still looks like glass. A clay marble weighs 9 grams.

What is the minimum amount of weighing needed such that you can identify the stack containing the clay ones?

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1 solution

With all the stacks lined up in a row, take 1 1 marble from the first stack, two from the second, and in general n n marbles from the n n th stack for 1 n 10 , 1 \le n \le 10, and then no marbles from the 11 11 th stack.

If all the stacks were composed of glass marbles then we would expect this set of 55 55 marbles to have a mass of 550 550 grams. If this is the measurement then we can conclude that the 11 11 th stack is composed of clay marbles. But if the n n th stack (for some n n such that 1 n 10 1 \le n \le 10 ) is composed of clay marbles, then the digital scale will read ( 550 n ) (550 - n) grams. Thus only 1 \boxed{1} measurement is required.

@Pi Han Goh Sorry about reporting your other question prematurely; after seeing that my report had been resolved, I re-read the question and realized that I didn't take your second condition fully into account. I still have one attempt left so I'll take my time.

I enjoyed this question, and I'm still trying to figure out a clever approach to your "near-narcissistic" question, (i.e., other than the brute-force approach I took).

Brian Charlesworth - 6 years ago

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Hint hint: BOUNDING!

Pi Han Goh - 6 years ago

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Yes, I did do some bounding, and that left me with a manageable number of cases, but it still didn't feel very clever. I'll give it a bit more thought .....

Brian Charlesworth - 6 years ago

Let X be the marbles of 10grams each. Let Y be the marbles with 9grams each. Take 1 marble from first stack,2 from second etc. Then we weigh. Let Z be the weight. Then 10X+9Y= Z (1). Now all marbles are 66 (1+2+3+...+11), meaning X+Y=66 =>X=66-Y (2). Substitute (2) at (1). 10(66-Y)+9Y=Z => Y= (660-Z)/9, remember Z is known cause we weighed the marbles we took out. For example if Y is 7, then the 7th stack contains the clay marbles.

John Miller - 5 years ago

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