Missing the Marks

Normally, a yardstick (36 inches long) has 35 marks placed on it at one-inch intervals, numbered 1 to 35. But suppose you are given a yardstick that only has the 1", 4", 9", 16", and 25" marks on it. Using only these marks and/or the ends of the stick, how many distinct lengths can be measured with this yardstick?

For example, a length of 7" can be measured from the 9" mark to the 16" mark, and a length of 11" can be measured from the 25" mark to one end of the stick.


The answer is 19.

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3 solutions

Sanjay Banerji
Dec 27, 2013

Actually I was thinking that we could measure all possible lengths ;

Example : if I need to draw a 10 inch line , will draw a 11 inch and then draw another 1 inch so that the latter is subtracted ...

7C2 -2 ...REASON:there are 7 marks that is, 0,4,9,16,25,36 taking combinations of any two we get the readings by their difference so 7 c2 but by inclusion exclusion we need to remove the common differences which are two in number .

Daniel Mathew - 7 years, 5 months ago

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You missed 1.

Ayush Pateria - 7 years, 5 months ago

Consider all numbers of the form x 2 y 2 x^2-y^2 , where x , y x,y are integers such that 6 x > y 0 , 6 \geq x > y \geq 0, . The number of different ordered pairs ( x , y ) (x,y) that fulfil the inequality are ( 7 2 ) = 21 {7 \choose 2}=21 . However, there exists two integers n n such that there exists two different pairs of integers ( x 1 , y 1 ) , ( x 2 , y 2 ) (x_1,y_1),(x_2,y_2) satisfying the above criteria where x 1 2 y 1 2 = x 2 2 y 2 2 = n x_1^2-y_1^2=x_2^2-y_2^2=n (these two numbers being 16 = 5 2 3 2 = 4 2 0 2 , 9 = 5 2 4 2 = 3 2 0 2 16=5^2-3^2=4^2-0^2, 9=5^2-4^2=3^2-0^2 ).

The total number of distinct lengths that can thus be taken are thus those that fulfil the above conditions, excluding repetitions, which gives us 21 2 = 19 21-2=\fbox{19}

Jared Low - 7 years, 5 months ago

This is what I thought initially as well, but then I realised that the question was asking which lengths can be measured without moving the ruler. Every number has a factor of one, so if we could move the ruler we could measure up to infinity inches.

Nicolas Bryenton - 7 years, 3 months ago
William Zhang
Dec 27, 2013

Count them:1,3,4,5,7,8,9,22,23,25,16,20,21,24,25,27,32,35,36

11,12,15 instead of the first 22, 23, 25

Sanjay Banerji - 7 years, 5 months ago

only 9 is repeating

Vighnesh Raut - 7 years, 2 months ago

A quick question: why isn't 0 included?

Carl Araya - 7 years, 5 months ago

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Cuz its mean that you dont measure anything. The problem ask "how many distinct length can be measured?"

Muhammad Ridwan Apriansyah B. - 7 years, 5 months ago
Pratik Vora
Dec 24, 2013

Consider marks from 0 to 36.From mark '0' 6 values are possible,from mark '1' 5 values,similarly from mark '25' 1 value.

Therefore total values=21.

Common values are 9(0-9,16-25) and 16(0-16,9-25).

So total values=21-2=19

I put 21 and got it wrong- but I tried seeing without trial and error- if there were any common values. Is there a smart way you can find the common values without trial and error?

Nahom Yemane - 7 years, 5 months ago

I just realized the two repeated ones...

Patrick Chen - 7 years, 5 months ago

JUST SIMPLY -THE ANSWER IS 7C2-2

Daniel Mathew - 7 years, 5 months ago

Oh right I just realised pythagoras

Nahom Yemane - 7 years, 5 months ago

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???

Sanjay Banerji - 7 years, 5 months ago

is there any method possible without trial and error to see the different possible cases, because for higher values this may be grotesquely tough

Sourav Chaudhuri - 7 years, 5 months ago

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