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April 16th, 2016, 06:32 AM   #1
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Physics Grad Stumped By A 12 Year Old's Math Problem

I am a little embarrassed that I have been unable to answer this question from my 12 year old son's Maths exam paper. I felt it was too tough for a 12 year old. Can anyone solve it?

Erdos runs a Maths Club. Each time they meet, they each drink a pint of coffee with milk. The amounts of coffee and milk may vary from cup to cup, but are never zero. Erdos drinks a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the club (including Erdos)? Make your reasoning clear.


It seems to present ONE equation, but with two unknown quantities. Conventional wisdom would therefore suggest the problem is impossible to solve.

Last edited by skipjack; April 16th, 2016 at 09:35 PM.
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April 16th, 2016, 07:46 AM   #2
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Math Focus: tetration
12N = 3m + 2c
(3m + 2c)/12 = a natural number

It is undetermined.

Last edited by skipjack; April 16th, 2016 at 07:56 AM.
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April 16th, 2016, 08:13 AM   #3
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Suppose they drink m pints of milk and c pints of coffee in total, where m > 0 and c > 0.

From what is stated about Erdos, m/4 + c/6 = 1.

Hence m = 4 - 2c/3 and so is less than 4.

If there are n other club members, n = (3/4)m + (5/6)c = (3/4)m + 5(1 - m/4) = 5 - m/2.

As n is a whole number, m is a multiple of 2.

As it's already known that m > 0 and m < 4, it follows that m = 2.

Hence n = 4, which means that the club has 5 members in total.
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April 16th, 2016, 11:20 AM   #4
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Just brilliant, Skip...I'm sure Iqbal will buy you an extra large coffee
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April 16th, 2016, 07:15 PM   #5
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I don't understand skipjack's solution. So I tried to verify his answer.

However, I am unable to do that either.

Would someone please write out a verification?

I'd probably fail that Grade 7 math exam.

Last edited by skipjack; April 16th, 2016 at 10:18 PM.
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April 16th, 2016, 10:27 PM   #6
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Quote:
Originally Posted by Timios View Post
I don't understand...
It is necessary to assume that 1 pint was the total of coffee and milk together that each drank, and that the overall totals ignore any wasted coffee or milk (or no wastage occurred), so that n + 1 = m + c, then subtracting 1 = m/4 + c/6 gives n = (3/4)m + (5/6)c, etc. Are there any later steps you didn't understand?

Last edited by skipjack; April 17th, 2016 at 12:44 PM.
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