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| The top nine at Solvemall Classic (Posted on 2010-06-01) |
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My pal is very fond of puzzles, but can't always remember all the details correctly. Yesterday he told me about a new version of the famous 9-hole golf problem.
‘OK, there’s this guy, and when he plays golf, he always uses one of 2 shots. He always aims straight for the hole; if he overshoots, then he plays back towards the hole with his next stroke…'
‘Yes, I get the idea. You’re saying that he has a drive of, say, x yards, and an approach shot of, say, y yards, some exact combination of x's and y's will always enable him to sink the ball. What are the lengths of the holes?’
‘Gosh, that’s the problem! I can only remember 4 for certain; 150 – that was the shortest - 200, 350, and the longest one was 500.’
‘And what do you ‘remember’ the rest were, old buddy?’
‘Hmmm, something like 180, 280, 300, 370, and 410 - but now I come to think about those numbers, I’m pretty certain that one or two of them are wrong.’
‘And how many strokes altogether?’
‘Hehe, that was what you had to work out – but none of the holes was less than 150 yards, and none took more than 5 shots.’
Can anyone tell me what the correct lengths of the 9 holes were, so that I can work out the original problem?
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Submitted by broll
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Rating: 4.0000 (1 votes)
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Solution:
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The starting point is to consider how many ways 2 different clubs can be used to generate the lengths of the holes (lengths). We assume y is bigger than x (if they were the same, many of the given values would be impossible – for the same reason, we can also be sure that the club values are relatively prime):
The possibilities are: x,2x,3x,4x,5x,y,2y,3y,4y,5y,x+y,2x+2y,2x+y,2x+3y,3x+y,3x+2y,y-x,y-2x,y-3x,y-4x,2y-x,2y-2x,2y-3x,3y-x,3y-3x,2x-y,3x-y,4x-y,3x-2y (33 solutions).
Since we do not know 2 of the solutions (and there is doubt about 2 of those we do have) we will not be able to find a unique solution unless we can find ways of reducing the number of possible candidates.
1. The problem says ‘combinations’ of x and y. If we omit those solutions that only have x or y, we save 10 solutions. (23 left)
2. If x and y are close in value, we can omit some of the solutions (e.g. if x is more than y/2, then y-2x will be negative). Since the minimum is 150, we can also omit solutions that would give a smaller value than that, or larger than 500.
Our pal suggests that he was able to ascertain that two of the values 180,280,300,370 and 410 were wrong, without knowing either the number of shots, or the accurate values. We note that every given value is a multiple of 10. This is therefore not a distinguishing feature; however it indicates that we can divide each length given, and the club values, by 10 without loss of generality. When we do that, we find that odd values of x and y will never sum to 12, 18, 24 or any other value of the form 6n. We can therefore infer that the values 180 and 300 are incorrect, which explains our pal’s comment. It follows that both club values are odd.
No doubt we were given the 'certain' values,150,200,350,500 for a reason. If we try combinations of odd numbers between, say, 5 and 19, we will find that some solutions eliminate themselves, e.g. we cannot ‘make’ 500 if the shorter club is 5 (=50)etc.
We are eventually left with 11 and 13 as the only club lengths that fit the profile of the ‘certain’ numbers and the other requirements of the problem.
The possible configurations are (ignoring those that are too small, too large, or already ruled out):
2*11+13= 35 (350) (given)
2*13-11= 15 (150) (given)
2*13+11 = 37 (370) (suggested)
3*11-13 = 20 (200) (given)
3*11+13 =46 (460) (?)
3*13-11= 28 (280) (suggested)
3*13-2*11 = 17 (170) (?)
3*13+11 =50 (500)(given)
4*11-13 =31 (310) (?)
4*13-11 =41 (410) (suggested)
There are 3 possibilites for the last 2 holes. Both 310 and 170 require 5 shots, while 460 requires 4 shots. But if we select 460 as one of the two, then we cannot be sure which of the others is right, while if we note that 170 = 180 – 10 and 310 = 300 + 10 we can select those two on the footing that though not minimizing the number of strokes, they share the property of being approximately right, and thereby reject 460.
So the original distances were 150,170,200,280,310,350,370,410,500, and the club lengths are 110 and 130.
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