Hardest Maths Question Ever
Maths and statistics discussion, revision, exam and homework help.
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n = number of students. n < 30.(Original post by Godsize)
Well, This is the hardest maths question I ever came across anyone who gets it right is a damn genious
A class has less than 30 students
Exactly 3/4 of them own footballs
Exactly 7/8 of them own football boots.
i)How many students are there in the class?
ii)How many students own both football boots and footballs?
3 in 4 own footballs, so 3n/4 is an integer.
7 in 8 own football boots, so 7n/8 is also an integer.
there are exactly 3 values on n for which both 3n/4 and 7n/8 are both integers:
n = 24, 16 and 8
Proof? Well, 3/4 = 6/8 and because the second fraction is 7/8, we can say that if there are s values for n where 7n/8 is an integer (and n is an integer), then there must also be s values for n where 6n/8 is an integer (and n is an integer). So we only need to worry about 7n/8.
So if n and 7n/8 are both integers, and n is less than 30, n must be a multiple of 8. If it is, then the division result will be in integer and this multiplied by 7 is also an integer. If it isn't, then the division result will not be an integer, and since 7 is prime, there are no non-integers which multiply by 7 to give integers. There are 3 multiples of 8 less than 30, since 8*4 = 32. They are 8, 16, and 24, whose factors with respect to 8 are 1, 2 and 3.
Hope this help, which it probably doesn't. The fact is there are 3 answers for the first problem. -
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Lol, well noted!(Original post by mik1a)
n = number of students. n < 30.
3 in 4 own footballs, so 3n/4 is an integer.
7 in 8 own football boots, so 7n/8 is also an integer.
there are exactly 3 values on n for which both 3n/4 and 7n/8 are both integers:
n = 24, 16 and 8
Proof? Well, 3/4 = 6/8 and because the second fraction is 7/8, we can say that if there are s values for n where 7n/8 is an integer (and n is an integer), then there must also be s values for n where 6n/8 is an integer (and n is an integer). So we only need to worry about 7n/8.
So if n and 7n/8 are both integers, and n is less than 30, n must be a multiple of 8. If it is, then the division result will be in integer and this multiplied by 7 is also an integer. If it isn't, then the division result will not be an integer, and since 7 is prime, there are no non-integers which multiply by 7 to give integers. There are 3 multiples of 8 less than 30, since 8*4 = 32. They are 8, 16, and 24, whose factors with respect to 8 are 1, 2 and 3.
Hope this help, which it probably doesn't. The fact is there are 3 answers for the first problem. -
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very true Xtina!! ... but the point im tryin to make is the question does not say ther are people with neither and the way that the questions sets out its parameters ensures ther is no children wiv neither, the question is askin for the single situation in which what they hav sed cud exsist properly, what u sed cud easily be applied to a class of 16 and 24 but you need a probability to determine which of the three class sizes, 8, 16 or 24, is the write one ...also the fact that it can be solved with the amount of info given wiv the gcse knowledge and gettin one answer with the way i posted makes me think thas wot ur ment to do...
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I have to agree with mak101's reasoning (but be careful with multiplying mate
). Using the approach with probabilities indeed leads to the answer i) 24 and ii) 15, provided, of course, that there are NO students who have nothing (if this was not so, the question would be impossible to solve).
but, mind you, if we forget about mak101's approach for a sec and if we look solely at the conditions of the problem (i.e. 3/4 owning ball, 7/8 owning boots), the answer could as well be 8 ppl in class, 5 owning both, 1 owning just a football and 2 owning just the boots, as this does not directly clash with the conditions given. so the problem is rather ambiguous and shouldn't have been set on an exam paper. -
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The solution is there is no solution.
Draw a square of area N (the number of people in the class).
Now draw a vertical line that divides the square into two regions having areas N/4 and 3N/4.
Now draw a horizontal line that divides the square into two regions having areas 1N/8 and 7N/8.
You should now have a square cut into four regions having areas N/32, 3N/32, 7N/32 and 21N/32. These correspond to the number of people having (a) no boots or ball, (b) ball but no boots, (c) boots but no ball and (d) both boots and ball.
Now since there is no integer N less than 32 that gives integer values for each of these proportions we can only conclude that there is no solution to the problem. -
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They are just different names. "Discrete" describes the nature of the maths, "decision" describes the nature of its applications. Different syllabi call it different things and cover different parts but it's the same strand of maths.(Original post by bono)
What is "D1" - Is it discrete, or decision, and is there a difference between these 2?
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I see.(Original post by XTinaA)
They are just different names. "Discrete" describes the nature of the maths, "decision" describes the nature of its applications. Different syllabi call it different things and cover different parts but it's the same strand of maths.
). Using the approach with probabilities indeed leads to the answer i) 24 and ii) 15, provided, of course, that there are NO students who have nothing (if this was not so, the question would be impossible to solve).
