# An Interesting Way to Combine Numbers

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1 An Interesting Way to Combine Numbers Joshua Zucker and Tom Davis November 28, 2007 Abstract This exercise can be used for middle school students and older. The original problem seems almost impossibly difficult, but there are obviously many ways to approach it by considering simpler problems. As they investigate it, students are actually drilling their multiplication and addition facts. In addition, depending on how detailed a study you want to make, the students will be forced to do a lot of simple algebra to obtain the results they need. They may also learn something about commutativity, associativity, and symmetric functions. 1 Introduction This document is meant for the teacher. It describes an interesting problem and then lists a number of ways the problem can be used in a classroom. Depending on the age and sophistication of the students, the classroom discussion can be taken in different directions. As usual, if you are the teacher, you will probably find this document more useful if, before reading our discussion about classroom presentation, you try to make some headway on the problem yourself. It will help you to think like a student and you may come up with additional ideas and strategies that did not occur to the authors. 2 The Problem Imagine that all the numbers from 1 to 100 inclusive are written on the blackboard. At every stage, you are allowed to erase two numbers that appear on the board (let s call the numbers you erased x and y) and in place of the two erased numbers, write the number x+y +xy. Repeat this operation until only a single number remains. What are the possible values for that remaining number? 3 Classroom Presentation Strategies Obviously you do not need to list all the numbers from 1 to 100. But write down a few of them, possibly something like this: 1, 2, 3, 4, 5, 6, 7, 8,..., 98, 99, 100 Also, depending on the sophistication and level of the class, you may not want to give the algebraic form x + y + xy; just say that the new number is obtained by adding three numbers: the two original numbers and their product. Do a couple of examples, as follows: 1

2 Suppose the two numbers you choose are 2 and 4, The new number is = 14. On the blackboard, physically erase the 2 and the 4 and write down the number 14. Point out that the board now has two copies of the number 14, and that there s nothing wrong with that. The two 14 s could even be combined at that point, yielding = 224. At every stage throughout your presentation, make sure the students do the calculations. Get the students to explain to you why the process must come to an end. (You want them to say something like At every stage two numbers are erased and only one number is added, so after each step there is one fewer number. Since we started with only 100 numbers, after 99 steps there will remain only a single number. ) This is a monotonicity argument : we can assign a positive number to a situation (in this case, simply the number of numbers remaining on the board) and show that at each stage this number is reduced (in this case, by one), and thus the process must eventually come to an end since once we are down to a single number the process must halt. Depending on the level of the class, you may need to work out one or two more examples of how the process works in general. You could show, for example, that if you begin with the numbers 1, 2, 3 and 4, that the process ends after three steps. Not only will this drive home the fact that all such processes have to end, but it will remind the students a few more times exactly how numbers are combined. If you work with {1, 2, 3, 4} as your initial set, don t choose pairs to eliminate in any straightforward order. For example, first combine 4 and 2 to yield 14. Then combine 1 and 3 to yield 7. Finally combine the 7 and 14 to yield the final result of 119. Ask the students if they have any ideas for how to proceed. Ask them if they can guess what strategies might lead to a large final number and which strategies might lead to a small final outcome. 4 How to Investigate the Problem It s pretty clear that it would be crazy to start work on the full problem that specifies that we begin with 100 different values on the blackboard. There are a tremendous number of ways that pairs of numbers can be selected and every one of them might yield a different final outcome. 4.1 A Smaller, Simpler Problem A good strategy in situations like this is to invent a similar, smaller problem that we can look at. We can look at problems that are both smaller (fewer than 100 initial numbers) or simpler. The word simpler can be a little misleading. Obviously we d like to try some examples with numbers that are easy to calculate with. This might mean small numbers, like 1, 2, 3, or perhaps look at the situation where all the initial numbers are the same (say 1, 1 and 1). What would happen if we allow the number 0 to be included? We can gain some insight into the problem with examinations like this, but later on we will also find that looking at the general algebraic form of the combining operation will actually be, in some senses, simpler, since it will show us how exactly how the original numbers fit into combinations, especially when we combine more than two of them. Ask the students to suggest simpler related problems, and you may get suggestions like, Just list the numbers from 1 to 10. That s a good suggestion, but try to convince them that even 10 is a very large 2

5 If b and c are combined first, we obtain b + c + bc and then we combine that with a to obtain: (b + c + bc) + a + (b + c + bc)a (3) = a + b + c + ab + ac + bc + abc (4) which is exactly the same result, and a symmetric expression in a, b and c as well! This means that if you have three numbers, the order in which you combine them makes no difference. The same calculation could be done for four numbers, et cetera, but again, the calculations start to become unmanageable. With four numbers it is not totally intractable, and if you wish to have the students do it, the result should be the following 15-term symmetric expression (where the initial list contains a, b, c and d): a + b + c + d + ab + ac + ad + bc + bd + cd + abc + abd + acd + bcd + abcd. A slightly sophisticated symmetry argument. In the calculations above, we calculated the value resulting from combining three numbers in two different orders. In fact, a slightly sophisticated symmetry argument shows us that this is true after only a single calculation. Equation 2 was calculated by combining first a and b, and then the result was combined with c. In the next line we recalcuated in a different order and obtained the same result, but that second calculation was unnecessary, in a sence, since the original result in Equation 2 is symmetric and there is no difference in the way that a, b and c enter into that calculation. Since the resulting equation is symmetric, that means that the inputs can be permuted any way you want, and in this case, the input basically specified the order of combination of initial numbers. Because of the symmetry, we can do the input in different orders and obtain the same result, so the second calculation to obtain Equation 4 was unnecessary. This argument may be too complicated for many students, but it is an important idea. At this point, we can take one more step toward generality that will simplify notation and perhaps improve comprehension. In the preceeding, we have repeatedly used expressions like combine a and b. In reality, it is just an arithmetic operation that doesn t have a common name or symbol. We will simply invent a new symbol, that represents our new operation. In the same way that a + b uses the symbol + to indicate the addition of a and b, we will use our new symbol, to indicate our combining operation. Symbolically, a b = a + b + ab. If you consider the calculations we did in equations 2 and 4, since a, b and c were arbitrary numbers, we have: (a b) c = (b c) a. (5) We also noted earlier that: a b = b a. (6) In other words, the operator is commutative. Because of this commutativity, we could rearrange the terms in equation 5 as follows (which puts it in the usual form used to express the result that the operation is associative): (a b) c = a (b c). (7) Equations 6 and 7 say that our new operation is both commutative and associative. Since it s commutative and associative, it doesn t matter in what order the numbers are selected to be combined; the result is always the same. 5

6 At this point, there are a couple of ways to go, depending on the sophistication of the class. The easy way is just to note that Associativity means that you don t need parentheses and Commutativity means that you can swap the order of the operands any way you want. With these two ideas, it s clear that any sequence that combines any number of initial numbers leads to the same result, and that may be good enough and you should skip the following section. 4.2 Commutativity and Associativity: Detailed Drill One problem students have with general formulas like: A (B C) = (A B) C or even simple formulas like the Pythagorean theorem: A 2 + B 2 = C 2 is that it is hard for them to get the idea that the A, B, and C can represent general expressions. For Pythagorean triples, it may be obvious to students that if you know that legs of a right triangle have lengths A = 3 and B = 4 that they can plug those numbers in to obtain a formula for C, but they don t know how to find C in a triangle where the lengths of the legs are given as A = 1 + u 2 and B = 1 u 2 like this: C 2 = (1 + u 2 ) 2 + (1 u 2 ) 2 = 2 + 2u 4, so C = 2 + 2u 4. To step through the details for rearranging formulas involving the operation using the primitive formulas for commutativity and associativity, students will have to understand and apply this idea over and over. If you are interested in drilling the ideas of commutativity and associativity, you can go through an example with combining, say, five numbers to show that commutativity and associativity implies that two different combining orders yield the same result, as follows: Suppose the five initial numbers are a, b, c, d and e. Your claim is that no matter how the numbers are combined, the result will be the same as combining d with e, then that number with c, then that number with b, and finally, that number with a. Suppose the students say this: Combine b with d (call that x). Next combine c with e (call that y). Now combine y with a (making z) and then z with x (producing the final number N). You probably want to give names to each step as the students make them up, like this: x = b d y = c e z = y a N = z x You can then work backwards to find a mathematical expression for N. Begin with the final equation above, and do one substitution at each step. In the list below we substitute first for x, then for z, then for 6

7 y to obtain each successive line: N = z x N = z (b d) N = (y a) (b d) N = ((c e) a) (b d) Make sure the students see that this final equation expresses in one line exactly the operations that they suggested. Next, we can take the final equation that involves only our original variables and convert it, step by step, to our target form: N = a (b (c (d e))). Here are the steps that will do it (see below for the detailed explanation for each step): N = ((c e) a) (b d) (8) N = (a (c e)) (b d) (9) N = a ((c e) (b d)) (10) N = a ((b d) (c e)) (11) N = a (b (d (c e))) (12) N = a (b ((c e) d)) (13) N = a (b (c (e d))) (14) N = a (b (c (d e))) (15) We will see that each equation is derived from the previous using only associativity and commutativity of the operation. We ll use the following shorthand. If we apply commutativity, we effectively convert X Y to Y X. This will be represented as in the second line below, where X stands for a and Y stands for (c e). If associativity is involved, we will convert A (B C) to (A B) C (or the reverse). The easiest one to see is in the seventh line below. Notice the strategy: we use commutativity to bring the letters into the order a, b, c, d, e, and associativity to group them the way we want. Equation 8: The original equation. Equation 9: Commutativity: X = a and Y = (c e) Equation 10: Associativity: A = a, B = (c e), C = (b d) Equation 11: Commutativity: X = (c e), Y = (b d) Equation 12: Associativity: A = b, B = d, C = (c e) Equation 13: Commutativity: X = d, Y = (c e) Equation 14: Associativity: A = c, B = e, C = d Equation 15: Commutativity: X = d, y = e 5 The Final Result Now that we know that the order in which we combine the original numbers on the board makes no difference, we can compute a few simple examples by using simple arithmetic. It s not too hard to do this 7

8 (and the class should) work out values for initial sets containing up to six natural numbers, {1}, {1, 2}, {1, 2, 3},..., {1, 2, 3, 4, 5, 6}. The following table lists the results they should obtain: Initial List Final Value 1 1 1, 2 5 1, 2, , 2, 3, , 2, 3, 4, , 2, 3, 4, 5, As you help with the calculations, you can point out a couple of things if the students themselves don t notice them. First, as you re working on the first two or three, point out that since the order doesn t matter, students can use whatever numbers they want that seem like they ll make for easier calculations. Then, say after the value 23 is computed for the third example, point out that rather than starting from scratch for the fourth example, note that they ve already combined 1, 2 and 3 and to obtain the next number, they just need to combine the 23 already obtained with 4 to obtain: = 119. Given this idea, the students may want to calculate a couple of more steps. (If they do, the answers for initial lists containing the first 7, 8 and 9 whole numbers are given by 40319, and , respectively.) Sophisticated students may from this information alone be able to guess the general formula, which, for the initial list {1, 2, 3,..., n} is given by (n + 1)! 1. Even if you ve guessed the formula, it still may be difficult to write down a convincing argument that it is correct. One approach is this: 5.1 Wishful Thinking When we see the formula x y = x + y + xy, we may notice that the form on the right can almost be factored. If only there were an additional 1 on the right, we d be in business, since (1 + x)(1 + y) = 1 + x + y + xy. Well, just add a 1! But you can t do that without changing the value of the expression, so you d better subtract a 1 to make up for it: x y = 1 + x + y + xy 1 = (1 + x)(1 + y) 1. As we shall see, this is a much better form of the algebraic expression that represents x y. If we use this form and want to calculate x y z, let s see what happens. (Note that because of associativity, we don t need to worry about parentheses in our expressions.) x y z = ((1 + x)(1 + y) 1) z = (1 + (1 + x)(1 + y) 1)(1 + z) 1 = (1 + x)(1 + y)(1 + z) 1 Point out where the pair 1 and 1 inside the initial set of parentheses in the second line above comes from that when we apply to two numbers, we add one to each, multiply them, and subtract 1 from the 8

9 result. One of the numbers is (1+x)(1+y) 1 and the other is z. Adding 1 to both gives (1+x)(1+y) and (1 + z), multiplying them gives (1 + x)(1 + y)(1 + z), and subtracting 1 gives the result. Lead the students through another couple of steps exactly as above to show that: x y z = (1 + x)(1 + y)(1 + z) 1 x y z w = ((1 + x)(1 + y)(1 + z)(1 + w) 1 x y z w v = ((1 + x)(1 + y)(1 + z)(1 + w)(1 + v) 1 It would be possible, of course, to write down a formal proof by mathematical induction that the general formula holds, but just doing the operation a few times should make the general pattern clear. The general pattern is that you add 1 to every number in the list, multiply all of those numbers together, and subtract one from the final result. Thus the result obtained from an initial list containing {1, 2, 3, 4, 5, 6, 7} is given by: (1 + 1) (2 + 1) (3 + 1) (4 + 1) (5 + 1) (6 + 1) (7 + 1) 1 which is: = 8! 1. Clearly, the result for an the initial list from 1 to 100 always yields 101! 1, which, in case anyone wants to multiply it out, is the following 160-digit number:

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