Five More Problems

When I was in eighth grade, one of our local newspapers would occasionally publish a math problem and wait for at least a week for someone to write in a solution.  I know better than to try that here, since the people I thought might read my last set of problems apparently didn’t.  Since the previous problems were more logic than math, I promised one actual math problem.  Three of the five here use math.  Since the last group was such a dud, I promise no more problems after this.

Problem 1:

Mary’s father has five daughters.  The first one he named Nana.  The next three were Nene, Nini, and Nono.  What did he name his fifth daughter?  (Tough problem, isn’t it?)

Problem 2:

In the following problem, each letter stands for a digit.  The same letter stands for the same digit, and different letters stand for different digits.  DO + DO + DO = GOO.  What are the corresponding digits?

Problem 3:

Sue leaves the house at the same time every day, driving the same route at exactly the speed limit, to pick up her husband at the train station precisely as the train arrives.  Speed limits to the station and back home are the same.  One day he takes the train that leaves an hour earlier, and begins walking home as soon as the train arrives.  Sue leaves at her normal time and picks him up on the way and turns around, arriving home 10 minutes early.  How many minutes did her husband walk?

Problem 4:

The hour hand on a 12-hour clock is between 9:00 and 10:00, while its minute hand is pointing in exactly the opposite direction, forming a line with the hour hand.  What time is it (hours, minutes, seconds, and fractions of a second)?

Problem 5:

Now it’s time to recall complex numbers.  For a complex number ‘z’, its norm (or absolute value) is written |z| and is the square root of the sum of the squares of its real and imaginary parts.  Complex numbers are plotted on a real plane with the real component along the x-axis and the imaginary component along the y-axis.

Show that for complex numbers u, v, and z, if u-qv is not zero, the equation |(z-u)/(z-v)|2 = q describes either a line or a circle.


Solution 1:

The fifth daughter’s name is Mary.  My favorite response to this answer is “Who’s Mary?”  Since Mary has a father, she must be his daughter.

Solution 2:

The equation can be rewritten 3 x DO = GOO.  We know that 3 x O must be 0+O, 10+O, or 20+O.  The only digits for O that match this are 3 x 0 = 0 and 3 x 5 = 15.  If O = 0, then 30 x D = 100 x G.  The only integers solving this are D=G=0, which is impossible if O is already set to 0; so O=5.  Now we have (30 x D) + 15 = (100 x G) + 55.  This gives us 30 x D = (100 x G) + 40, or 3 x D = 10 x G + 4.  3 x D can only end in 4, when D=8 and G=2.

The answer is 85 + 85 + 85 = 255.

Solution 3:

What makes this problem so difficult is that it doesn’t appear to give enough information, but it’s simpler than it seems.  If the time it takes Sue to drive to the train station is ‘t’, then the time required for the round trip is ‘2t’.  When her husband was walking, Sue’s round-trip time was 2t – 10 minutes, so it took half that time (t – 5 minutes) to reach him, or 5 minutes less than the normal time to get to the station.

Since he had started walking an hour earlier than normal arrival time, it took him (60 – 5) minutes to reach Sue.  He had been walking for 55 minutes.

Solution 4:

Let’s begin by trying to solve the exact time in revolutions of the hour and minute hands.  The minute hand makes 12 revolutions for every revolution of the hour hand, so it travels 12 times faster.  If ‘h’ is the fraction of a revolution where the hour hand stands, and ‘m’ is the fraction of a revolution where the minute hand stands, then m = 12h – 9.  (This is the same as the 12 times the fraction of a revolution past 9:00.)  In addition to this, the hands are pointing in opposite directions, so in terms of revolutions, h – m = ½ revolution.

Since m = h – ½, h – ½ = 12h – 9, giving 11h = 8½ or h = 17/22.  Converting revolutions back to hours, we get 12h = (12 x 17) / 22, or time in hours = 102/11 = 9 + 3/11.  Now we have 3/11 hours = 180/11 minutes = 16 + 4/11 minutes.  Next we have 4/11 minutes = 240/11 seconds = 21 + 9/11 seconds.

So, the exact time is 9:16:21 + 9/11 seconds.

Solution 5:

Break the complex numbers down to their real coordinates.  Take u = a + ib, v = c + id, and z = x + iy.  Then q = |((x-a) + i(y-b)) / ((x-c) + i(y-d))|2 = ((x-a)2 + (y-b)2) / ((x-c)2 + (y-d)2).

Multiplying and gathering terms, we get (1-q)x2 – 2(a-qc)x – 2(b-qd)y + (1-q)y2 = q(c2 + d2) – (a2 + b2).

If q=1, the equation is linear in x and y and describes a line.  If q is any other value, then (x – (a-qc)/(1-q))2 + (y – (b-qd)/(1-q))2 = ((q(c2+d2) – (a2+b2))/(1-q) + ((a-qc)2 + (b-qd)2 + q2((a-c)2 + (b-d)2) / (1-q)2, which is a circle with respect to x and y.  The only tough part in this last problem is making sure you don’t screw up the algebra.


I tried to make this challenging, even if you’ve had honors algebra.  The last three problems are much more difficult than the ones from when I taught the comparable college courses.  They are the first rung on the ladder if you expect to compete in either the Putnam exams or the International Mathematics Olympiad.

Speaking of honors courses, we didn’t have those when I went to school, but we had something much better.  I had an interesting teacher in Junior High and an outstanding one in High School.

Please let me know whether these problems were completely out of your ballpark or if they were relatively easy (or somewhere in between).


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