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The Man-Made Universe
By Sherman K. Stein
Dover Publications, Inc.Copyright © 1999 Sherman K. Stein
All rights reserved.
Questions on Weighing
This chapter will raise some important questions about numbers. While they may seem to be mere recreational puzzles that could be understood and investigated by anyone who can do simple arithmetic, in fact they concern a fundamental property in number theory. Not until Chapter 3 will the "why" behind the answers be considered. The goal of this chapter is to offer an opportunity to experiment with an open-ended problem and to see how a single mathematical idea can appear in a variety of disguises.
The questions, which concern weighing, will be introduced by a few examples. Say that we have a two-pan scale of the type seen in chemistry labs and statues of "Justice":
Furthermore, we have an unlimited supply of 5- and 7-ounce measures. Now, supposing that potatoes weigh only whole numbers of ounces, rather than any amount as they actually do, let us ask which potatoes we would be able to weigh with our balance and our two types of measures.
For instance, using only the 5-ounce measures, we can weigh 5, 10, 15, 20, 25, 30, 35, ... ounces. Or using only the 7-ounce measures, we can weigh 7, 14, 21, 28, 35, ... ounces. Moreover, we could place one of each type together on a pan:
Thus we can weigh 12 ounces, 12 = 5 + 7. Or we could put one of each type of weight alone on a pan:
In this way we can weigh 2 ounces, since a potato of this weight, together with the 5-ounce measure, balances the 7-ounce measure.
Can we weigh a 3-ounce potato? Yes, by placing two 5-ounce measures on one pan and a 7-ounce measure with the potato:
The balancing records the equation 3 + 7 = 2 · 5.
Can we weigh a 4-ounce potato? Yes. For instance, by placing two 5-ounce measures with the potato and two 7-ounce measures on the other pan:
The corresponding equation is
4 + 2 · 5=2 · 7.
Or we could place three 7-ounce measures with the potato, which then balance five 5-ounce measures. The equation in this case is 4 + 3 · 7 = 5 · 5.
Can we weigh a 1-ounce potato? Even this can be done, as the reader may prefer to work out for himself before reading the next sentence. Two 7-ounce measures and the potato on one pan balance three 5-ounce measures on the other pan. The reader may wish to pause and devise still other ways of measuring this 1-ounce potato with 5- and 7-ounce measures.
Once we know that we can weigh a 1-ounce potato, then we know that we can weigh any number of ounces. For instance, we can weigh a 6-ounce potato as follows. First recall that two 7-ounce measures and a 1-ounce potato balance three 5-ounce measures:
Repeating this arrangement of measures six-fold weighs a six-ounce potato. That is, from the relation 1 + 2 · 7 = 3 · 5 5 it follows that 6(1 + 2 · 7) = 6(3 · 5) or
6 + 12 · 7 = 18 · 5.
Of course, this may not be the simplest method for weighing a 6-ounce potato. Indeed, 6 + 3 · 5 5 = 3 · 7, so we could have managed by placing three 5-ounce measures with the potato and three 7-ounce measures on the other pan. But the reasoning at least assures us that if we can weigh a 1-ounce potato with a supply of two types of measures, then we can weigh any whole number of ounces with those measures.
So much for the combination 5 and 7. Suppose we turn to another combination, 8 and 21. Even if we have only 8-ounce and 21-ounce measures available, we can measure a 1-ounce potato, since
1 + 3 · 21 = 8 · 8.
Eight 8-ounce measures on one pan will balance the potato and three 21-ounce measures on the other pan.
The reader may now suspect that perhaps any pair of measures can weigh a 1-ounce potato. But this is not so. If, for example, we have only 6-ounce and 8-ounce measures, then we could never hope to measure a 1-ounce potato, or, for that matter, any odd number of ounces. (The reader should pause to think about why this is so.)
We are now in a position to ask some basic questions. Suppose we have at our disposal an unlimited supply of measures of two types. How can we decide whether we can weigh a 1-ounce potato with them? For instance, can we use 539-ounce and 1619-ounce measures to weigh a 1-ounce potato? More generally we can ask: What potatoes can we weigh with an unlimited supply of two given types of measures? Keep in mind that all potatoes and measures weigh a whole number of ounces (0, 1, 2, 3, 4, ...). The whole numbers will usually be referred to as the natural numbers.
The questions really concern numbers, not potatoes. Let us gradually translate the second question into the language of numbers: Denote the weights of the two measures by A and B ounces respectively. In our first combination we had A = 5 and B = 7. The weight of the potato will be denoted by W ounces.
There are various methods of weighing the potato. One consists in putting several A-ounce measures on the pan with the potato and several B ounce measures on the other pan. How many of each we use will depend on W, A, B, and our arithmetic. Say that we use X of the A ounce measures and Y of the B -ounce measures:
The corresponding equation is
W + XA = YB,
an equation that asserts merely that the scale in the figure balances. (We omit the multiplication sign between letters.)
For A = 5 and B = 7, let us see what X and Y are for various W. When W = 1, for instance, we have 1 + 4 · 5 = 3 · 7. Here X = 4 and Y = 3. Also 1 + 11 · 5 = 8 · 7, so that X = 11 and Y = 8 also perform the weighing when W=1.
Another method of weighing consists in placing only B-ounce measures with the potato and only A-ounce measures on the other pan. For A = 5, B = 7, the equation 1 + 2 · 7 = 3 · 5 illustrates this. We have W + XB = YA, where W = 1, X = 2, Y = 3.
A third method consists in placing the potato on one pan and the measuring weights on the other pan. For example, if W = 12 we have 12 = 5 + 7; if W = 27 we have 27 = 4 · 5 + 1 · 7. This method corresponds to an equation of the type W = XA + YB.
No other practical method exists, for there would be no point in placing measures of equal weight on both pans, since they could be removed without affecting the balance. Thus we need to consider only three types of equations:
W + XA = YB, W + XB = YA, W = XA + YB.
In all of these, X and Y are to be natural numbers, possibly including zero. The question about potatoes now becomes one about natural numbers: Let A and B be natural numbers. For which natural numbers W can we find natural numbers X and Y such that at least one of these equations holds:
W + XA = YB, W + XB = YA, W = XA + YB?
The potatoes are gone, but we are left with three equations to deal with. We can simplify matters further (reducing the three equations to one) by making use of the negative numbers, -1, -2, -3, -4, -5, ..., which lie to the left of 0 on the number line. (See Appendix A.) The numbers
..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...
are called integers. (Note that any natural number is an integer.)
With the aid of the negative numbers, we will reduce the first two equations, W + XA = YB and W + XB = YA, to the third type, W = XA + YB. Consider first, W + XA = YB. This can be rewritten as W = (-X) A + YB, which is of the third type. (For instance, we may rewrite 1 + 4 · 5 = 3 · 7 as 1= (-4)5 + 3 · 7.) Similarly, W + XB = YA, where X and Y are natural numbers, can be reduced to the third form by writing it as W = YA + (-X)B.
The question about potatoes now reduces to this: Let A and B be natural numbers. Which natural numbers W can be expressed in the form W = MA + NB for certain integers M and N?
Let us see which values of M and N describe our earlier work for the combination A = 5 and B = 7. The following table records some cases we considered, beginning with 5, 10, ...; 7, 14,....
The row for "3", for instance, tells us
Just as a potato may be weighed in more than one way, so M and N are not necessarily unique. For the case W = 6 there were the two weighings
6 + 12 · 7 = 18 · 5 (6 = 18 · 5 + (-12)7)
3 · 5 + 6 = 3 · 7 7 (6 = (-3)5 + 3 · 7).
A little arithmetic produces still more representations of 6; for instance,
6 = 4 · 5 + (-2)7; 6 = 11 · 5 + (-7)7; 6= (-10)5 + 8 · 7.
Our first question, concerning the possibility of weighing a 1-ounce potato, now reads: For which pairs of natural numbers A and B can we find integers M and N such that 1 = MA + NB?
Take the combination A = 24, B = 73. Since 3 · 24 = 72, we have 1 = (-3)24 + 1 · 73. Thus for the combination 24 and 73 M and N exist. But what about the combination 24 and 75? Can they measure 1? What about 21 and 34? What about 89 and 233? And, we may wonder: If a pair of measures can't measure a 1-ounce potato, what is the smallest positive weight they can measure?
There is another way of looking at these questions. Recall the representation 1 = 8 · 8 + (-3)21. The basis of this is that 8 - 8 differs from 3 · 21 by just 1. We may have found this representation by listing 1 · 8 = 8, 2 · 8 = 16, 3 · 8 = 24, 4 - 8 = 32, 5 - 8 = 40, 6 · 8 = 48, 7 · 8 = 56, 8 · 8 = 64, ... and 1 · 21 = 21, 2 · 21 = 42, 3 ? 21 = 63, ... until we found an entry in one list that differed by 1 from an entry in the other list.
The numbers of the form "integer times 8" are called the multiples of 8. The multiples of 21, or of any integer, are defined similarly. This illustration shows some of the multiples of 8 and some of the multiples of 21:
Thus the question "Can A and B measure 1?" can be phrased in terms of multiples: "Are there multiples of A and of B that differ by exactly 1?" Since the multiples of a number are regularly spaced on the number line, the question can also be interpreted geometrically: If we have an unmarked ruler A inches long, and another B inches long, can we measure a distance of one inch?
The reader has probably guessed the answers to these general questions. Chapter 3 will return to this topic. There the equation W = MA + NB will turn out to be the basic tool for establishing fundamental properties of the natural numbers.
1. Is every natural number also an integer? Is 0 a positive integer? Is 0 a negative integer? Is every positive integer also a natural number?
2. Drawing pictures of the two-pan scale, show three different methods of weighing a 1-ounce potato with 4- and 7-ounce measures.
3. Drawing pictures of the two-pan scale, show three different methods of weighing a 1-ounce potato with 8- and 13-ounce measures.
4. What do you think is the smallest weight of a potato that can be measured with (a) 8- and 12-ounce measures, (b) 8- and 11-ounce measures, (c) 9- and 12-ounce measures? Why do you think so?
5. A 4-ounce potato and five 7-ounce measures balance three 13-ounce measures. Express this in the form (a) W + XA = YB,. where all letters stand for natural numbers, (b) W = MA + NB, where M and N are integers.
6. Are there integers M and N such that (a) 1 = M · 7 + N · 10, (b) 1 = M · 8 8 + N · 10, (c) 1= M · 13 + N · 22, (d) 1= M · 6 + N · 21? If your answer is "yes," give values of M and N; if "no," explain why.
7. a. Show how to weigh a 1-ounce potato with 11-ounce and 18-ounce measures.
b. Use (a) to find a method of weighing a 3-ounce potato with 11-ounce and 18-ounce measures.
8. Draw the balanced loading of the two-pan scale that corresponds to each of these equations: (a) 3 + 6 · 7 = 5 · 9, (b) 5 = 3 · 11 + (-4)7, (c) 23 = 2 · 4 + 3 · 5.
9. Can a 1-ounce potato be weighed with 5- and 8-ounce measures in such a way that (a) the potato is with the 5's, and the 8's are in the other pan; (b) the potato is with the 8's, and the 5's are in the other pan?
10. Can a collection of 5-ounce measures on one pan ever balance a collection of 8-ounce measures on the other pan?
11. The products 5 · 10 and 3 -17 differ by 1. Draw four different balanced loadings of the scales for weighing a 1-ounce potato that reflect this arithmetic fact. (The measures may be 5, 10, 3, or 17.)
12. Find integers M and N, if you think they exist, such that
a. I = M · 4 + N · 9,
b. 1 = M · 9 + N · 11,
c. 1 = M · 10 + N · 15,
d. 1 = M · 23 + N · 25.
13. a. Draw on a number line several multiples of 7 and several multiples of 12.
b. Do the multiples of 7 meet the multiples of 12 anywhere other than at 0?
c. What is the closest that a multiple of 7 can be to a multiple of 12 without actually coinciding with it?
14. Consider 5- and 12-ounce measures. Show that it is possible without using more than four 12-ounce measures to weigh (a) a 1-ounce potato, (b) a 2-ounce potato, (c) a 3-ounce potato, (d) a 4-ounce potato, (e) a 5-ounce potato, (f) a 6-ounce potato, (g) a 7-ounce potato. Will this be true for heavier potatoes?
15. a. Find a multiple of 9 that differs from a multiple of 11 by 1.
b. Show them on the number line.
16. List (a) four multiples of 3, (b) five multiples of 2, (c) six multiples of 7.
17. a. Show that 1 can be weighed with 8's and 11's.
b. Give some small values of W that can be weighed with 16's and 22's.
c. Give some small values of W that can be weighed with 24's and 33's.
18. Find four pairs of integers M and N such that (a) 12 = M · 2 + N 5, (b) 5 = M · 2 + N · 5, (c) 13 = M · 2 + N · 5.
19. What potatoes can be weighed with 4- and 7-ounce measures, if we never use more than three 7-ounce measures?
20. Consider weighings by 3- and 7-ounce measures. Draw a picture of balanced scales recording the equality 3 · 7 = 7 · 3. (a) Show that if a potato can be weighed by placing it on the pan with 3-ounce measures it can also be weighed by placing it on the pan with 7-ounce measures. (b) Is this true more generally than for just 3's and 7's?
21. Find integers M and N such that 1 = M · 5 + N · 9 and (a) M is positive, N negative; (b) M is negative, N positive. (c) Draw the corresponding loadings of the scales.
22. Find integers M and N such that 13 = M · 4 + N · 5 and (a) M and N are positive; (b) M is positive, N negative; (c) M is negative, N positive. (d) Draw the corresponding loadings of the scales.
23. a. What amounts of postage can you make with 5-cent stamps and 8-cent stamps? 2.
b. What type of weighing problem would be equivalent to the postage problem raised in (a)?
c. What requirement does the postage problem impose on M and N if the postage, W, equals M · 5 + N · 8?
24. What amounts of postage can be made with 4-cent and 7-cent stamps?
25. What amounts of postage can be made with 10-cent and 13-cent stamps?
26. What amounts of postage can be made if you are allowed to use at most one 1-cent stamp, at most one 2-cent stamp, at most one 4-cent stamp, at most one 8-cent stamp, at most one 16-cent stamp, and so on (each denomination being a product of 2's)?
Excerpted from Mathematics by Sherman K. Stein. Copyright © 1999 Sherman K. Stein. Excerpted by permission of Dover Publications, Inc..
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