Math 3250/5250 - Solutions 2

Fall 2001

Remember that the goal of these problems is not so much to get the right “answer,” but to understand and describe the solution process. Your analysis of the process is most important. If you don’t use the four-step Polya method explicitly, you should at least have it working in the background.

1. Reading the problem carefully is critical here. Drawing a few pictures also helps. You need to imagine the world consisting of “even-shakers” and “odd-shakers” (those who have shaken an even number of hands and those who have shaken and odd number of hands). The problem asks you to prove that the number of odd-shakers is even. I will give two solutions:

Solution 1: Imagine that no one in the world has yet shaken a hand. After the first handshake, there are two odd-shakers, which is an even number of odd shakers. From now on, as people begin shaking hands, one of three things can happen: (i) if two even-shakers shake hands, they both become odd-shakers, which increases the number of odd shakers by 2, so there is still an even number of odd shakers; (ii) if an odd-shaker shakes hands with an even shaker, the number of odd-shakers remains unchanged and even; (iii) if two odd-shakers shake hands, the number of odd-shakers decreases by 2, so there is still an even number of odd shakers. In all cases, the number of odd-shakers remains even for any number of handshakes.

Solution 2:

· If two people shake hands, we count it as two handshakes - one for each person.

· It follows that regardless of how many people are considered or how they shake hands with each other, the total number of handshakes is even (a multiple of 2).

· The total number of handshakes is also the sum of all handshakes by the even-shakers and all handshakes by the odd-shakers. Thus,

· The sum of all handshakes by the even shakers is always even (because they have all shaken an even number of hands).

· Thus, the sum of all handshakes by the odd-shakers must also be even.

· But we are not done! Does it follow that the number of odd-shakers is also even? Yes: the sum of all handshakes by the odd-shakers can be even only if there is an even number of odd-shakers (the sum of a bunch of odd numbers is even only if there is an even number of numbers).

If you have studied graph theory, this result is the analog of the theorem that the number of vertices with odd degree in any graph is even.

2. Visualize the situation (or draw a picture). Let x be the quantity of antifreeze (in liters) that is removed and then replaced with the 90% solution. After the replacement, the radiator has x liters of 90% anti-freeze and (21 - x) liters of 18% anti-freeze. (Remember this means percent by volume.) The required solution has a concentration of 42%, which means it has 0.42 × 21 liters = 8.82 liters of antifreeze. Therefore the amount of 90% anti-freeze that must be added satisfies

0.90 x + 0.18 (21 - x) = 8.82

Solving for x, we find that x = 7 liters of 90% anti-freeze must be added. You should check that 7 liters of 90% solution and 14 liters of 18% solution results in 21 liters of 42% solution.

3. I can’t imagine solving this problem without a well-labeled picture. The other key point is that the time required to complete each of the two legs is the same for both boats. Recall that time elapsed = distance/rate. Let the (constant) speeds of the boats be U and V. Let x be the (unknown) width of the river. Referring to the picture below, if we equate the times for both boats to travel the first leg, we find that

Equating the times for the second leg, we find that

We have two equations in three unknowns. However, notice that the ratio U/V can be eliminated leaving a single equation for x. Solving this equation, we find that the width of the river is x = 1700 yards.

4. Every word in this conversation is a clue! First, list all the possible ages of the children such that the product of the ages is 36.

1, 1, 36 → sum = 38 1, 2, 18 → sum = 21 1, 3, 12 → sum = 16 1, 4, 9 → sum = 14 1, 6, 6 → sum = 13 2, 2, 9 → sum = 13 2, 3, 6 → sum = 11 3, 3, 4 → sum = 10

The fact that the sum of the ages equals today’s date provides two clues. First, the first age combination (1, 1, 38) can be eliminated because a date cannot be greater than 31. But notice that all the sums are different except those for (1, 1, 6) and (2, 2, 9). So if the ages were any of the other combinations, Paul would be able to figure out the ages. Because he can’t figure out the ages, only the combinations (1, 1, 6) and (2, 2, 9) remain. The last clue, “the oldest has red hair,” says that there is an oldest child, so (1, 6, 6) is eliminated. The children have ages (2, 2, 9).