PRIZE PUZZLE FOR OCTOBER 2006
FOUR LITTLE SWEETS
What was the cost of the items?
15, 20, 24, 30.
I received 4 correct answers (and none incorrect). They were from Michael Hodge, Clem Robertson (who won the draw), Colin Rowe and John Stafford. Well done and thank you.
How to do it.
This is an unusual problem and I would not like to answer a problem of this kind with much larger figures. Here is how I reasoned to get the answer (though as I composed the problem I knew in advance what the answer was!).
The four percentage ratios are 6:5, 5:4, 3:2 & 2:1.
There being four items and four ratios between them implies some redundancy so we first look for two ratios that equal a third and notice that 5/4 * 6/5 = 3/2. This is likely but not necessarily the ratios that relate three of the items but any other arrangement does not work. For example, if these three (call them A, B, C) were such that B = 5A/4 and C = 6A/5 we have a minimum value for A of 20 (LCM of 4 & 5) and a sum of 20 + 25 + 24 = 69 requiring another 20 for the fourth item which is not a ratio 2:1 of any of the others. Any multiple of these three is too big.
If B = 5A/4 and C = 6B/5 so that C = 3A/2 we get A = 4, B = 5, C = 6, plus multiples. The sum of these three is 15 with multiples 30, 45, 60, 75 leaving 74, 59, 44, 29, 14 for the fourth item respectively, none of which have a ratio of 2:1 with any of the others.
Another possibility is B = 6A/5 with C = 5B/4 = 3A/2 making A = 10 as the smallest. We then have the lowest three values as 10, 12, 15 with a sum of 37 leaving 42 for the fourth which does not work. Doubling A,B,C makes the sum 74 leaving 15 for the fourth and this is indeed a ratio 2:1 (or 1:2) of 30. The answer is therefore 20p, 24p, 30p and 15p, a total of 89p with ratios 6/5 = 24/20; 5/4 = 30/24; 30/20 = 3/2 and 30/15 = 2/1.
This was not an easy problem but I have to stretch the abilities of the usual band of solvers from time to time. Apologies to the rest!