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.
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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 .
Consider all numbers of the form x 2 − y 2 , where x , y are integers such that 6 ≥ x > y ≥ 0 , . The number of different ordered pairs ( x , y ) that fulfil the inequality are ( 2 7 ) = 2 1 . However, there exists two integers n such that there exists two different pairs of integers ( 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 (these two numbers being 1 6 = 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 2 1 − 2 = 1 9
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.
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
only 9 is repeating
A quick question: why isn't 0 included?
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Cuz its mean that you dont measure anything. The problem ask "how many distinct length can be measured?"
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?
I just realized the two repeated ones...
JUST SIMPLY -THE ANSWER IS 7C2-2
Oh right I just realised pythagoras
is there any method possible without trial and error to see the different possible cases, because for higher values this may be grotesquely tough
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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 ...