1
Morrison
Olivia Morrison
Prof. Winkler
Writing Assignment #3
05.0902.2013
The “Smaller Numbers” Method: A More Manageable Puzzle
The “smaller numbers” method of mathematical puzzle solving is a techniquefocused on finding a pattern that leads to the solution. In a suitable puzzle, the reader will pick numbers smaller than those given in the puzzle and substitute them in. He or she will then attempt to solve the puzzle using these smaller numbers. If done correctly, the pattern of a solution should hopefully emerge. If it does indeedbecome apparent, the reader can then apply this pattern to the puzzle using the larger, given number. The pattern should yield the correct solution, even if finding such an answer seemed impossible at first. The “smaller numbers” technique can thus be used to find a pattern and eventually arrive at the solution to a mathematical puzzle by breaking it down to a much more manageable size.
The puzzle-solving method of testing smaller numbers to reach the ultimate solution is a common one, and is a strategy that directs the reader to a (usually generalizable) pattern. Using smaller numbers in succession – for example, groups of three, four, and then five people if the problem is discussing a group of eight people – can unearth a general pattern. The reader can then use this pattern as direction towards the final solution.The best way to learn how to recognize a problem in which the smaller numbers strategy can be used is by seeing examples of such problems. A simple example of this type of puzzle is as follows:
In recent years a number of clever coin-weighing or ball-weighing problems have aroused widespread interest. Here is a new and charmingly simple variation. You have 10 stacks of coins, each consisting of 10 half-dollars [see figure 9.5]. One entire stack is counterfeit, but you do not know which one. You do know the weight of a genuine half-dollar and you are also told that each counterfeit coin weighs one gram more than it should. You may weigh the coins on a pointer scale. What is the smallest number of weighings necessary to determine which stack is counterfeit? (Gardner 232)
Instead of trying to solve this problem in its original form immediately, we can break it down to smaller numbers. How many weighings will it take to determine which of three stacks is counterfeit? Four stacks? Five? For three stacks of coins, we can figure out which stack of coins is counterfeit in one weigh-in. We know the difference in weight between the normal and counterfeit stacks (1 gram), and we know the weight of a real coin. Let us call that weight x grams. Then, if we take one coin from the first stack, two from the second, and three from the third and weigh all six coins at once, we can see which stack is counterfeit. If we only took one coin from each stack, we would be able to tell that there was a counterfeit coin being weighed, but not to which stack it belonged. By taking a different amount of coins from each stack, we can differentiate between the three stacks. In this case, if the weight is 6x+1 grams, it is the first stack because there is a one-gram difference. We can set up the weights as:
(x)+2(x)+3(x)=6x
So, if the weight increases by one, the weight formula becomes:
(x+1)+2(x)+3(x)=5x+(x+1)=6x+1
Similarly, if it is the second stack, the formula will be:
(x)+2(x+1)+3(x)=4x+2(x+1)=4x+(2x+2)=6x+2
And for three, it is:
(x)+2(x)+3(x+1)=3x+3(x+1)=3x+(3x+3)=6x+3
We can apply the same strategy if there are four stacks of coins, simply by taking four coins from the fourth stack and adding them to the six taken from the first three stacks. Then, the four options of weights become 10x+1, 10x+2, 10x+3, and 10x+4 grams. The same goes for five stacks as well.After adding five half-dollars from the fifth stack to the combined pile,we will have constrained the total weight to 15x+1, 15x+2, 15x+3, 15x+4, or 15x+5 grams. We can then extrapolate this information to form a general pattern. For ten stacks of coins, we take one coin from the first stack, two from the second, three from the third, and continue in this manner until we take ten coins from the tenth stack. So, we have 1+2+3+…+10=55 coins. The weight differential will correspond to the counterfeit stack. If all the stacks were of a normal weight, the combined selected coins would have a weight of (55)x grams. Since there is a difference of one gram per counterfeit coin, the increase in weight from the baseline (55)x grams equals the number of the counterfeit stack (1 gram = first stack, 2 grams = second stack, …, 10 grams = tenth stack). Gardner’s solution follows this formula:
The counterfeit stack can be identified by a single weighing of coins. You take one coin from the first stack, two from the second, three from the third, and so on to the entire 10 coins of the tenth stack. You then weigh the whole sample collection on the pointer scale. The excess weight of this collection, in number of grams, corresponds to the number of the counterfeit stack. For example, if the group of coins weighs seven grams more than it should, then the counterfeit stack must be the seventh one, from which you took seven coins (each weighing one gram more than a genuine half-dollar). Even if there had been an eleventh stack of 10 coins, the procedure just described would still work, for no excess weight would indicate that the one remaining stack was counterfeit. (246)
Thus, the “smaller numbers” method allows us to develop a pattern to solve the puzzle. That pattern can then be applied to the specific numbers mentioned in the puzzle itself to reach a final solution. In this case, we went from three stacks, to four, to five, and eventually reached the specified 10. From there, we came up with a direct solution to the puzzle, using the pattern derived from the smaller numbers.
The strategy of using smaller numbers to reach the overarching solution is a method of puzzle solving that can be applied to almost any puzzles using specific numbers and a strategy. The solution strategy, usually, is a type of pattern, and will continue to work if the scenario is broken down into smaller numbers as well. Thus, if the reader is able to recognize this type of mathematical puzzle, he or she can utilize the “smaller numbers” method as a manner in which to attempt a solution. The best way to employ this method is by taking much smaller numbers – such as 3, 4, 5, etc., depending on the number(s) used in the actual puzzle – and substituting them for the larger numbers in the puzzle. Then, if the reader is able to solve the puzzle with these smaller numbers, he or she should hopefully be able to determine a pattern that will work when used in combination with the puzzle’s original, larger numbers. Thus, by originally using the smaller numbers, the reader has broken the puzzle down into a smaller, more manageable form. He or she is then able to arrive at the correct solution when he or she might not have otherwise thought such an answer to be possible.
As demonstrated in the counterfeit coins puzzle, the reader can apply the pattern found by using smaller numbers to the original values. However, oftentimes the pattern does not end with the given number. There are many cases in which the pattern continues to n, and can thus be applied to any number of the reader’s choosing. An apt example of the type of puzzle containing this continuous pattern is as follows:
Bob and his wife Alice attend a party with five other couples. When they arrive, each person greets the rest and shakes the hand of some number of people at the party, never shaking his or her spouse’s hand. At the end of the evening, Bob asks everyone there, including Alice, how many hands they shook. He receives eleven different answers. How many hands did Bob shake at the beginning of the night? (Winkler)
Once again, we see that the original puzzle may prove difficult to solve. With eleven different answers and twelve people, it just seems that there is too much information – and indeed too much vague information – to solve this puzzle. We can try to find a solution by using the “smaller numbers” method. We start with only two couples: Bob and Alice and their two friends. Again, Bob asks Alice and the two others – let’s call them Charlie and Dottie – how many hands they shook at the beginning of the evening. This time, he receives three different answers: one for each of the people he asked. First, we must examine the possible answers. Because no one shakes his or her spouse’s hand, the maximum number of hands one person could have shaken is two: the hands of the other couple. Because there are three distinct possible answers, the remaining two must be one and zero. Now we see who shook how many hands. A diagram will help:
We picked Dottie to have shaken two hands, so she must have shaken the hands of both Bob and Alice. Alice, then, shook one hand: Dottie’s. We know that Charlie shook no hands, and from the diagram, we can tell that Bob shook one hand (Dottie’s). Another notable feature of this diagram is that the person who shook the most hands (Dottie) is married to the person who shook no hands (Charlie). We can repeat the process using three couples, and keeping these new ideas in mind. The maximum number of hands shaken becomes four, and the number of different answers Bob receives is now five. The diagram is as follows:
Here, we see that, again, the person who shook the most hands is married to the person who shook no hands. We can also note that the person who shook the next-highest amount of hands is married to the person who shook the next lowest. And both Bob and Alice shook two hands. Thus, the number of hands shaken between the couple adds up to the total possible number of handshakes (4+0, 3+1, 2+2). If this pattern holds, we can apply it to the original case of eleven total couples. There, we should see that the person who shook ten hands is married to the person who shook none (10+0), the person who shook nine to the person who shook one (9+1), and so on, until we reach Bob and Alice, each of whom must have shaken the same number of hands, according to our pattern. Thus, each must have shaken five hands (5+5): our final answer.
We can extrapolate this pattern and apply it to a case in which there are n+1 couples. The maximum number of hands shaken in this case is n. So the person who shook n hands is married to the person who shook zero hands (n+0=n), the person who shook n-1 hands to the person who shook one ([n-1]+1=n), and eventually we reach Bob and Alice. Each must have shaken n/2 hands: (n/2)+(n/2)=n. Thus, the pattern can be generalized, and will remain applicable no matter the number of couples.
The “smaller numbers” approach to finding the solution to a mathematical puzzle is exceedingly helpful due to its focus on finding a pattern. Once the reader is able to solve the puzzle using smaller numbers in place of the given number, he or she will hopefully be able to recognize a pattern in the method of solving the puzzle. When he or she eventually picks up on this pattern, it can be applied to the original numbers given in the puzzle. Hopefully, this will yield the correct solution to the puzzle, and thus the method will have been successful. In many cases, the pattern is generalizable, in addition to being applicable to the original puzzle, and thus can be used with other numbers in place of the original as well. The reader can recognize puzzles where this “smaller numbers” method might be employed by recalling similar puzzles where it has been successful. Generally, if there is a strategy/pattern involved in solving the puzzle and a large number – whether it be 100 men, ten stacks of coins, or six couples – is a key component, the reader can utilize the “smaller numbers” method, or some variation of it. The “smaller numbers” method, then, is an incredibly useful means by which a reader can arrive at the solution to a complicated puzzle.
Works Cited
Gardner, Martin. The Colossal Book of Short Puzzles and Problems: Combinatorics, Probability, Algebra, Geometry, Topology, Chess, Logic, Cryptarithms, Wordplay, Physics, and Other Topics of Recreational Mathematics. Ed. Dana Richards. New York: Norton, 2006. Print.
Winkler, Peter. Math 7: Mathematical Puzzles. Dartmouth College, Hanover, New Hampshire. 2 Apr. 2013. Class Lecture.