Puzzles

Puzzle 76

I took a logic class in college that met five days a week, Monday through Friday, and one Monday, at the very beginning of class, the teacher announced, “In one class meeting this week, I will give you a surprise quiz.” Since this was a logic class, we were all able to figure out which was the only day he could give us the quiz. Which day was it? (It’s not enough to show why your answer works; you must also show why each of the other four possible days does not work.)

Show Solution

Since Friday is the last day of the week that he could give us the quiz, if he didn’t give us the quiz by Thursday, then we would know at the end of that class meeting that he would have to give us the quiz on Friday, so it wouldn’t be a surprise and, therefore, he couldn’t give us the quiz on Friday. That meant that the last day of the week that he could give us a surprise quiz was Thursday, so if he didn’t give us the quiz by Wednesday, we would know at the end of that class that he would have to give us the quiz on Thursday, so again it wouldn’t be a surprise and, therefore, he couldn’t give us the quiz on Thursday. By the same reasoning we could determine that he could not give us a surprise quiz on Wednesday or on Tuesday. This leaves Monday as the only day of the week that he could give us a surprise quiz, that is, a quiz that we didn’t know about before the class meeting in which the quiz was to be given.

Puzzle 77

The numbers, 3, 4, 5, 6 and 7, are typed and printed onto five distinct cards, arranged in an addition column, and the same is true for the numbers, 8, 7, 6, 3 and 2. Exchange one card from the first column with one card from the second column in such a way as to get the same sum in both columns.

Show Solution

If you are methodical, you would first ascertain that the first sum is 25 and the second is 26 and if the difference between them is an odd number, there is no way to meet the conditions of the puzzle without some trick being involved. The solution is to take the card with the “6” on it from the first column, rotate it 180 degrees so the “6” becomes a “9”, and exchange it with the card with an “8” on it in the second column, so both sums become 27.

Puzzle 78

This one is for high school geometry students. Without using Heron’s Formula, which is usually not taught in a basic geometry course, anyway, find the area of a triangle whose sides measure 13, 14 and 15.

Show Solution

The formula for the area of a triangle is $A=\frac{1}{2}bh$ , where b is the measure of the base and h is the height, i.e., the measure of the altitude drawn to that base. Draw the altitude of the triangle to the side measuring 14, thereby creating two right triangles with a common altitude. Call the length of this altitude h and the measures of the bases of the two right triangles x and 14 – x. Applying the Pythagorean Theorem to each of the two right triangles, we get $x ^{2}+h ^{2}=13 ^{2}$ and $(14-x)^{2}+h ^{2}=15 ^{2}$ . Solving this system of simultaneous equations, we get x = 5 and 14 – x = 9, and one more application of the Pythagorean Theorem yields h = 12, so the area of the original triangle is 84.

Puzzle 79

You are in a race and you overtake the person who is in first place. What place are you in, now? OK, so you got that one right away. How about this one: You are in a race and you overtake the person who is in last place. What place are you in, now?

Show Solution & Corollary Puzzle

The person who is in last place is behind everyone else, so you can’t be behind the person who is in last place, so you can’t overtake the person who is in last place. OK, so you are in a race and the person who is in last place overtakes you. What place are you in now?

Show Solution to Corollary

Not enough information. If you are in first place when you are overtaken, you are then in second place. If you are in sixth place when you are overtaken, you are then in seventh place. A common error is to say that you are in last place after being overtaken by the person who was formerly in last place, but that would be true only if you are in the next to last place when you are overtaken.

Puzzle 80

Find unequal rational numbers, a and b (other than 2 and 4), such that $a^{b}= b^{a}$ .

Show Solution

Since a and b are both rational, let b = ra, where r is rational. Then $a ^{ra}=(ra) ^{a}$ , $a ^{r}=ra$ , $a ^{r-1}=r$ , $a = {r^{\frac{1}{{r - 1}}}}$ . Since r is rational, a is rational whenever $\frac{1}{r-1}$ is an integer, i.e., whenever r – 1 is the reciprocal of an integer, so $r-1=\frac{1}{k}$ , where k is an integer, so $r=1+\frac{1}{k}=\frac{k+1}{k}$ . Therefore, $a=(\frac{k+1}{k}) ^{k}$ and $b=ra=(\frac{k+1}{k})(\frac{k+1}{k}) ^{k}=(\frac{k+1}{k}) ^{k+1}$ . These values for a and b satisfy the given equation for k equal to any non-zero integer, so there are an infinitude of solutions. For example, for k = 2, $a=(\frac{2+1}{{2}}) ^{2}=\frac{9}{4}$ and $b=(\frac{2+1}{{2}}) ^{2+1}=\frac{27}{8}$ , so ${a^b} = {(\frac{9}{4})^{\frac{{27}}{8}}} = {[{(\frac{3}{2})^2}]^{\frac{{27}}{8}}} = {(\frac{3}{2})^{\frac{{27}}{4}}}$ $= {[{(\frac{3}{2})^3}]^{\frac{9}{4}}} = {(\frac{{27}}{8})^{\frac{9}{4}}} = {b^a}$ . And in general, if $a=(\frac{k+1}{k}) ^{k}$ and $b=(\frac{k+1}{{k}}) ^{k+1}$ , where k is a non-zero integer, then a and b are both rational, and ${a^b} = {[{(\frac{{k + 1}}{k})^k}]^{{{(\frac{{k + 1}}{k})}^{k + 1}}}} = {(\frac{{k + 1}}{k})^{k(\frac{{k + 1}}{k}){{(\frac{{k + 1}}{k})}^k}}}$ $= {(\frac{{k + 1}}{k})^{(k + 1){{(\frac{{k + 1}}{k})}^k}}} = {[{(\frac{{k + 1}}{k})^{k + 1}}]^{{{(\frac{{k + 1}}{k})}^k}}} = {b^a}$