Answer to Riddle #75: NSA Track & Field
So this puzzle was written by a chap called Steven working for the NSA. They publish a puzzle or riddle every so often here, the previous puzzles have been a bit too verbose for me, it doesn't mean you wont like them. This is the September 2016 puzzle:
Puzzles like this tend not to be directly solvable in the sense that we can't write down a series of equations and just turn the handle until the answer falls out. What we have to do is start to build the model and see what works and what doesn't, usually something stated in the question will make more sense when we get to the point where we an use it. For example we could try X, Y & Z being 3, 2 & 1. (It wont be that simple, this is a hard NSA puzzle.) Then take a look at the points Adam, Bob, and Charlie have and see if we can find a model that fits. It turns out the best place to start is with there being 40 points awarded overall.
You should check out the assumptions at the bottom of the page.
Number of Events and Points Per EventThere are 40 points overall (22 + 9 + 9) which means the number of points per event multiplied by the number of events must be forty. We now need to look at the different combinations of points per event and number of events and see what we can eliminate with what else we know. For example there could be 1 Event with 40 Points Per Event, except that can be the solution because we know there are at least two events:
- 1E x 40PPE - Not possible we know there are at least 2 events
- 2E x 20PPE - Not possible. We know Adam and Bob each won an event, with only 2 events there is no way Bob & Charlie could finish on the same number of points.
- 4E x 10PPE - Viable at this stage
- 5E x 8PPE - Viable at this stage
- 8E x 5PPE - given the condition x > y > z > 0 and that they are all integers, the lowest possible values are 3, 2, 1 which means we have a minimum number of Points Per Event of 6.
- 10E x 4PP - as above
- 20E x 2PP - as above
- 40E x 1PP - as above
If we are hypothesising 10PPE we are X + Y + Z = 10. Keeping in mind that 'x > y > z > 0, and all point values being integers.'
- 5,4,1 - Since our hypothesis is 4 events, if the maximum number of points available per player is 5 there is no way for Adam to get to 22. The greatest available would be 20
- 5,3,2 - As above
- 6,3,1 - 6 + 1 + 1 + 1 = 9 which is good for Bob and Charlie separately but actually it's not possible for them both to finish last in 3 events. OR there is no way to make 22 points out of 6,3&1 over 4 events
- 7,2,1 - We can make 22 and 9 points so this is viable for now.
For 8 Points Per Event there are only two combinations that fulfil the criteria 'x > y > z > 0, and all point values being integers.' They are 4,3,1 and 5,2,1.
Points BreakdownWe can eliminate 4,3,1 because Adam scored 22 points. Since we know there were 5 events the maximum he could have scored would have been 5 x 4 = 20.
We have now proven by elimination there 5 events with points awarded 5, 2 & 1 for 1st, 2nd and third respectively. But we still haven't answered the question...
Who gets the points-For Adam to get 22 points he must have finished 1st in 4 events and second in 1 other. We're told that Bob won the Javelin so that fits.
-Bob finished first in the Javelin 5 points and so must have finished last in the other 4 events. 4E x 1 point. Makes 9 points.
-Charlie must have finished last in the Javelin and second in every other event to get 1 + 2 + 2 + 2 + 2 = 9 points.
So Charlie finished second in the 100m dash.
AssumptionsWe're assuming no ties and no eliminations. Traditionally in the event of a tie each person gets to share the higher place. So if first is a tie both first places get the points of the winner, there is no second and the next highest person is placed 3rd.
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