An Old Ruler
Chris has an old 6-inch ruler that is missing the 4 in mark. However, he can still measure all integer lengths from 1 in to 6 in . He'd like to see if he can erase more marks and still be able to measure all integer lengths from 1 in to 6 in .
What is the minimum number of marks the ruler could have?
Chris' ruler can measure all integer lengths from
1
in
to
6
in
with only 4 marks.
Note that all measurements must be made directly using the ends of the ruler and/or its marks. For example, you can't repeatedly use a distance of 1 in to make larger distances.
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8 solutions
Another possible answer may be combination of 1 and 4 mark.
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Yes, that is true, but the problem states that the ruler is missing the 4-in mark.
its possible for any such case other than two extremes with a difference of 3 as "(i) 1 and 4 (ii) 2 and 5
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Actually, no. As already mentioned, to measure 5, there needs to be a marking at either 1 or 5. However, markings at 1 and 5 can't measure 2 or 3, markings at 3 and 5 (or by symmetry 1 and 3) can't measure 4, and markings at 4 and 5 (or by symmetry 1 and 2) can't measure 3. This leaves the only possible markings at 1 and 4 or at 2 and 5.
Grt one!!!
I like this explanation a lot because it has pictures and words for different types of learners.
Wait, there is a 6CM? I didn't know
How did you make picture? Just Paint?
To measure 6 different lengths, you need at least 6 pairs of marks (with each end counting as a mark). So you want the smallest n such that n choose 2 is greater than or equal to 6. That n is 4; subtracting the two ends, you need two marks on the ruler.
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If you had a 10-inch ruler, then by this strategy you want the smallest n such that n choose 2 is greater or equal to 10, and that n is 5, and subtracting 2 for the ends gives 3, for 3 marks on the ruler. But I believe you need more than 3 marks to measure the 10 different lengths on a 10-inch ruler.
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I said you need "at least" n choose 2 marks. i did not claim that that number of marks would be sufficient (although I could have expressed myself better on that point), just that fewer marks could not work at all. Finally, I was dealing specifically with the six-length case, not the general case. I expect that a better minimum could be found for the general case, and that such a minimum will depend on the number of lengths needed: the more lengths you need to measure, the more marks will be needed.
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@Michael Sommers – I wonder if a general formula can be found for a ruler of n inches to give the exact minimum of marks needed to measure all the integers.
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@David Vreken – I think there probably is one. I also think there is a pattern that can be followed in assigning the marks. With an n-unit ruler, you start with the ends at 0 and n. Then you need 1 and n - 2. This gives you lengths of 1, 2, n-3, n-2, n-1, and n, which is all you need for n = 6. For n = 7, you need one more mark to get you the 3. Put it at 4. For n = 8, you now have marks at 0, 1, 4, n-2 = 6, and 8, which gives you all the numbers. I"m too lazy to carry on, but you get the idea.
There must be at least one mark that is one away from one of the ends. However, this mark is not sufficient to measure 2, 3, or 4, so at least one more mark must be made and the only possible solution with two marks by hit and trial . That's what I did!
1 more guy did this problem in same way by the way
How do you measure 4
If you only have the 1 in mark you can measure all integers as well. You just need to use the ruler more then once, which is not excluded in the question.
Moderator note:
It is stated in the instructions
Note that all measurements must be made directly using the ends of the ruler and/or its marks. For example, you can't repeatedly use a distance of 1 in to make larger distances.
This was my thought as well. Poorly worded question.
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I thought the same as well.
No, it is not.
Yes, by flipping the ruler around and measuring from the end where there could be a 6, you need no extra markings.
This is addressed in the note at the end.
Question was unclear.
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Agreed. If you read the question literally, "What is the minimum number of marks the ruler could have?", the answer is, "Zero". It doesn't specify the goal of the number of marks at all, so there could be a rule with no marks and it wouldn't appear to be violating anything with this question.
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"He'd like to see if he can erase more marks and still be able to measure all integer lengths from 1 in to 6 in." "Note that all measurements must be made directly using the ends of the ruler and/or its marks. For example, you can't repeatedly use a distance of 1 in to make larger distances." Sure, if you just ignore the entire context of the problem and the image attached, you would be right. But the problem isn't just the single bolded question there. It's the entirety of the information provided along with that bolded question. If you are trying to answer that question without looking at the problem, then that is entirely your fault for intentionally ignoring the problem.
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@Zanthor Nolastname – I interpret it as: "Some of the marks have completely worn away, including the 4in mark. So we know it still has at least how many marks?"
No, it is not.
And how will you know where to put the ruler after taking it off? You'll need to make a mark for reference somewhere and the question was clear about "Note that all measurements must be made directly using the ends of the ruler and/or its marks."
PS: sorry for my english
You also cant assume that the length of the ruler is exactly 6 inches. Every ruler I've ever used has been longer than it's intended measure.
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I mean, we can assume the length of the ruler is exactly 6 inches, if we're talking theoretical mathematics for the sole purpose of solving a logic puzzle.
I've sometimes seen rulers that are exactly the length they are made to measure up to. This must be such a ruler in order to achieve the minimum number of marks.
By the labels shown in the introduction to the problem, not only can you assume that the ruler is exactly six inches, you must assume this.
It is stated that you can't just repeat one over and over again so two is right
If we have only one marking a , then we can measure at most three distances: a , 6 − a , 6 .
If we have two markings a , b , a < b , then we can measure distances a , b , b − a , 6 − a , 6 − b , 6 . We need to choose a and b such that these distances are all distinct. We can immediately notice that neither a nor b is 3 since we would then have a = 6 − a or b = 6 − b . But, for the same reason 6 − a and 6 − b are not equal to 3 either. Thus, the only option is b − a = 3 .
Now, we have distances a , a + 3 , 3 , 6 − a , 3 − a with condition that a < 3 , i.e. a = 1 or a = 2 . You can easily check that both cases give solution.
Thus, the minimum number of markings is 2 and it can be achieved in two ways: ( 1 , 4 ) and ( 2 , 5 ) .
Remark. It's obvious that there is even number of solutions by symmetry, just measure from the opposite side of the ruler. This corresponds to transformation x ↦ 6 − x .
(1,4) is not valid since the 4 mark is already removed.
I solved this by making tables of numbers that is needed to be measured from 6, 5, 4, 3, 2, 1. Then I started to figure out the requirements that length can be measured. Then I figured out that 6 can be ignored as that is the total length of the ruler meaning that it can be always measured.
- 5 can be only measured if you have 1st or 5th in marks on the ruler.
- 4 can be only measured if you have 2nd mark on the ruler.
- 3 can always measured if you have both 2nd and 5th between them.
- 2 can always measured if you have 2nd mark on the ruler.
- 1 can be always measured if you have 5th in mark on the ruler.
Then I deduced from the facts that I have that I don't need 1st mark to able to measure all lengths. So the answer is 2nd and 5th marks.
It says that 4 cannot be a mark
I think there should be two marks next to each other to see the scale.
the mark of the ruler is 1 and 4.
In order to measure 5, there must be at least one mark that is one away from one of the ends. However, this mark is not sufficient to measure 2, 3, or 4, so at least one more mark must be made. One possible solution with two marks is shown below, so this is the minimum number of marks needed to measure all integer lengths 1, 2, 3, 4, 5, and 6.